A test team of Gaoyi Chemical verifies the composition of a certain magnesium aluminum alloy

From the conservation of charge, the following ratio is obtained:

1 mol of aluminum can produce mol of hydrogen, 1 mol of magnesium can produce 1 mol of hydrogen, and the first part of hydrogen gas, i.e., mol, can be obtained to contain mol of magnesium, then the total sample contains mol of magnesium.

The second part produces mol of hydrogen, i.e., mol, and if it contains aluminum x mol, then there is:

x = x = 1 15 mol The total aluminium content in the sample is 2 15 mol.

The mass of aluminum is 27 x (2 15) =

The ratio of the amount of aluminum-magnesium is (2 15): =4:9

The first step of the reaction generates magnesium metaaluminate, aluminum does not participate in the reaction, I don't know if this step is right, I calculate it according to this. If you are wrong, there is a problem with this step, and this answer is for reference only and gives you an idea.

In the future, you can try to solve the problem with conservation of charge, as long as you know the reactants and products, which is faster and there is no need to write chemical equations.

There should be a problem with the original question, and the first one should be a reaction with sodium hydroxide.

Only aluminum reacts with sodium hydroxide solution, and both aluminum and magnesium react with hydrochloric acid, and according to the conservation of electrons, the first hydrogen gas is mol, that is, it contains aluminum, and the two parts contain aluminum. The second part produces hydrogen, i.e., it can be concluded that it contains magnesium, then the total sample contains magnesium mol.

The mass of aluminum is, and the mass of mg is 24*

The ratio of the amount of aluminum-magnesium substances is::1

Aluminum and sodium hydroxide produce sodium metaaluminate and hydrogen, and magnesium does not react; hydrochloric acid reacts with magnesium and aluminum to produce hydrogen; Now you should be able to do it. Hydrogen is aluminum, hydrogen is produced by aluminum, and the rest is produced by magnesium, 1mol of magnesium, which seems very simple.

2H2O+2Al+2Naoh=2Naalo2+3H2 gas symbol, then the reaction between magnesium and aluminum does itself.

The answer is aluminum, magnesium.

In fact, this problem only requires the amount of al, and does not need to use the condition of the volume of hydrogen generated.

The addition of sulfuric acid dissolves all the old sheds of the alloy, so the metal becomes the corresponding sulfate, and the sulfuric acid is reduced to produce hydrogen. al~ h2, mg~h2。There is no mention of "just completely dissolved", so there may be an excess of sulfuric acid.

Sodium hydroxide is added, and NaOH reacts with possible excess sulfuric acid on the one hand, and metal ions on the other. The clever point here is that if you imagine that sodium hydroxide causes all metals to produce corresponding hydroxide precipitates and neutralizes excess sulfuric acid, then the amount of sodium hydroxide and sulfuric acid is equivalent (here the amount of hydrogen ions and hydroxide ions is the same, and the ratio of the two substances is 2:1).

Further, the Al(Oh)3 actually generated will react further with sodium hydroxide, Al(Oh)3+OH-=AlO2 - 2H2O. Because of this reaction, the amount of sodium hydroxide exceeds the amount of sulfuric acid by 2 times when the precipitated mass happens to not change. According to the equation described by Shangyou, it can be known that the extra sodium hydroxide and aluminum hydroxide are pressed by 1:

1. React.

After analysis, the idea of finding the amount of Al is very clear. Let the amount of its substance be a.

The amount of hydroxide ions (from sodium hydroxide) is calculated to be more than hydrogen ions (from sulfuric acid), and the excess amount is all used to react with Al(OH)3.

There is conservation, al al(oh)3 alo2 - oh- (excess).

Therefore, the amount of OH- beyond H+ is equal to the amount of AL.

a= mol•l－1)*1 - 01l*( mol•l－1)*2= mol

In the above equation, the term after the minus sign and finally multiplied by 2 represent that one sulfuric acid contains two hydrogen ions.

2al + 2naoh + 2h2o = 2naalo2 + 3h2；

2al +6hcl = 2alcl3 + 3h2；mg + 2hcl = mgcl2 + h2；

Let al: Al be Duan Li xmol; mg is ymol; Take the envy.

；x = mol；

y = = ；y = ；

1) m(al) = =

Moore's Equation:

mg+2hcl=mgcl2+h2

2al+2naoh+2h2o==2naalo2+3h2x=

2al+6hcl=2alcl3+3h2

mg+2hcl=mgcl2+h2

y---y== the ratio of the amount of matter :1

1) Let the quantities of Mg and Al be x and y respectively, then:

mg+2h+

fe2+h2

x x2al+6h+

2al3+3h2

Y can be obtained according to the quality of the two Wang imitation mill: xx 24g mol + y 27g mol=, x+, according to the trapped bucket solution: x =, answer: the amount of magnesium in this large reed alloy is;

2) The mass of Al in the alloy is:, and the mass fraction of Al in the alloy is:

Answer: The mass fraction of Al in the alloy is.

Let's take a look at the diagram first, O to A is because the aluminum magnesium ions react with sodium hydroxide to form the corresponding hydroxide precipitation, A to B is that the aluminum hydroxide continues to react with sodium hydroxide to form sodium metaaluminate, and B to C is when the NaOH is completely excessive, the aluminum is completely dissolved, and the rest is all magnesium hydroxide, which can be seen from the level of the BC segment, NaOH

According to the comparison of the experimental data of A and B, it can be seen that the mass of gold in combination B increases compared with the mass of gold in combination A, and the mass of hydrogen gas generated increases, and 30 ml of hydrochloric acid solution of the same concentration is taken in each of the three groups, indicating that the hydrochloric acid in group A is excessive, and a and d are incorrect;

According to the comparison of the experimental data of B and C, it can be seen that the mass of gold in combination C increases compared with that of gold in combination B, and the mass of hydrogen gas generated increases, and 30 ml of hydrochloric acid solution of the same concentration is taken in each of the three groups, indicating that the hydrochloric acid in group B is excessive, and B and D are incorrect.

Combination B gold increases mass compared to group A gold: Combination C gold increases mass compared to group B gold.

The increase in mass of hydrogen generated by combination B gold compared with the hydrogen generated by combination A gold: the increase in mass of hydrogen generated by combination C gold compared with hydrogen generation by combination B gold = (:(1>51:64, indicating that there is excess gold in combination C and insufficient hydrochloric acid, and c is correct.

The same hydrochloric acid solution was taken from each of the three groups of equal mass; Comparing the data of the two groups A and B, it can be seen that adding more alloys produces more gas, indicating that the hydrochloric acid of group A is sufficient; Comparing the data of the two groups of ethylene and propylene, it can be seen that adding more alloys produces more gas, indicating that the hydrochloric acid of group B is sufficient; Comparing the data of the two groups, it can be seen that the mass of the alloy that produces gas consumption is , and the mass of the alloy that produces gas consumption is 0,385g, indicating that there is a surplus of alloys and the amount of hydrochloric acid in group C is insufficient

Therefore, C

1) Group A: Calculated according to the chemical equation, 2AL+2NAOH+2H2

o=2naalo2

h2, magnesium aluminum alloy is all the mass of aluminum, the amount of the substance is, the amount of sodium hydroxide substance is needed, so the volume of sodium hydroxide solution =

According to the principle of choosing a graduated cylinder, it is large and close, and a 100ml graduated cylinder should be selected, so the answer is: 100;

If the solid surface is stained with sodium metaaluminate solution, the weight of the remaining solid will be large, the mass fraction of the measured aluminum will be reduced, and the measured mass fraction will be low; The method to prove that the solid is washed clean is to verify the existence of sodium metaaluminate solution design experimental verification, the experimental design is: take a small amount of washing solution drop by drop to add dilute hydrochloric acid to excess, if there is no precipitation in the whole process, it means that the washing is clean;

So the answer is: low; Take a small amount of washing solution and add dilute hydrochloric acid drop by drop to excessive, if there is no precipitation in the whole process, it means that the washing is clean;

2) Group B. 1) The analysis process and steps can be known to determine the mass fraction of the alloy by measuring the volume of gas, and the device needs to have intact air tightness, and the air tightness of the device should be checked; The inspection method is: add a certain amount of water to C, then increase C so that the liquid level is higher than B, let it stand for a while, if the liquid level does not change, the air tightness is good;

So the answer is: check the air tightness; Add a certain amount of water to C, then increase C so that the liquid level is higher than B, let it stand for a while, if the liquid level does not change, the air tightness is good;

2) Make the liquid level in B and C equal, and then need to cool the temperature of the solution to room temperature and read the gas volume at eye level, so as to avoid the error of the hot solution on the reading;

Therefore, the answer is: whether the gas is cooled to room temperature before measuring the gas with the measuring tube, and the head-up reading;

3) The analysis of the experimental process shows that the starting trachea is 112ml, and a sufficient amount of hydrochloric acid, magnesium and aluminum are added to the reaction to generate hydrogen, and the volume of the trachea is 448ml, indicating that the hydrogen generated is 336ml; The other end is added to the sodium hydroxide solution and aluminum to react to generate hydrogen, and the final trachea reading is 672ml, indicating that the volume of hydrogen generated is 224ml, and the mass fraction of aluminum is calculated according to the chemical equation, and the amount of magnesium in the alloy is x, and the amount of aluminum is y:

mg+2hcl=mgcl2

h2，1 1

x x2al+6hcl=2alcl3

3h2，2 3

y2al+2naoh+2h2

o=2naalo2

3h2，2 3

yx+vmx=

y=3vmmol

Mass fraction of aluminium =

3vmvm100%=60%, so there is no need to convert the gas volume to the volume of the standard condition, so the answer is: no; 60%．