The first year of junior high school uses the simple calculation of chemical equations to be anxious

1. 2p + 5o2 = 2p2o5

x62/222) = (

x = the mass of water is 100 16*18= the mass of hydrogen is 100 16*2=

p --p2o5

m(p2o5)=

Write the mass corresponding to the relative molar mass under the equation) 2H2O=O2+2H2

m(h2o)=100kg*2*18(g/mol)/32(g/mol)=

m(h2) = according to the law of conservation of mass).

The mass of water is 100 16*18= The mass of hydrogen is 100 16*2=

(1) Cu-Zn alloy because Zn can react with dilute sulfuric acid. The mass score is then calculated. However, the composition of Cu-AG alloy products does not react with dilute sulfuric acid.

2）zn+h2so4=znso4+h2

x y 20+

So x=y=the mass fraction of copper is (

Mass fraction of solute.

It can be seen that the mass of hydrogen produced m(h2)=

fe+h2so4==feso4+h2

56 152 2m(fe) m(feso4)

m(fe) = m(feso4)=, so the mass fraction of iron in the pig iron sample is::6=

The mass fraction of the solute in the solution obtained after the reaction is:

1: The quality of hydrogen generated in the second time can be judged by the quality of the zinc that is removed in the second reaction.

zn + 2hcl == zncl2 +h2 ↑

x solution yields x=2In the end, even if hydrochloric acid is added, the solid mass will not decrease, indicating that there is no substance left that can react with hydrochloric acid, (all that remains is copper). So the rest is copper, so.

Copper mass fraction =

1.The mass of zn participating in the reaction for the second time is. Let the mass of hydrogen produced be x.

zn + 2hcl *****=zncl2 + h265 2

x65:2=:x x=So the mass of hydrogen produced for the second time is .

2.According to the law of conservation of mass, the mass of zn of the first reaction is also , so the mass of the sample is . Due to the excess hydrochloric acid, the remaining solids are all copper, ie.

The mass fraction of copper in the brass sample is The result is to calculate it yourself with a calculator, I don't have it next to me)

Solution 1, quantity difference method.

Let the mass of copper oxide in the original mixture be x

H2 + Cuo = H2O + Cu mass reduction.

x80/16=x/ x=4g 4/5=80%

Solution 2 Chemical formula method.

H2 + Cuo = H2O + Cu The reduction of solids in the reaction is the mass of oxygen in Cuo i.e. 5 g = g

Mass fraction of oxygen in cuo = (16 80)=4 g, 4 5=80%.

Solution 3, let the mass of copper oxide in the original mixture be x, then the mass of copper in the original mixture is 5g-x, and the mass of copper generated by reaction is.

h2 + cuo = h2o + cu

xx=4 g 4 5=80% Answered!

1.(1) Let the mass of hydrogen peroxide be x.

2h2o2 = 2h2o + o2

x 64gx=136g

2）136g/10%=1360g

A: ......2.(1) Let the mass of zinc be x and the mass of sulfuric acid y y.

zn + h2so4 = znso4 + h2

x yx=y=(2) requires a 10% sulfuric acid solution:

A: ......Note: Both questions are ninth-grade questions, very basic calculation questions, why can't they? Read the textbook carefully, think carefully and comprehend.

1》First, the reaction equation is: 2 H2O2 = 2H2O + O2 mol mass of oxygen: 32g mol

So n(o2) = 64g 32g mol

2 mol so h2o2 =4 mol, so m=4x34=136g136g=m x 10%.

1360g2》 The reaction equation is: Zn + H2SO4 = ZnSO4 + H2H2 = H2

The reaction is one to one, so n(zn) = mol

So m(zn) = * 65 = g

From the above reaction, we can see that N(H2SO4) = mol

So m(h2so4) = * 98

So = m*10%.

m = 98g

The limestone taken is the mass g

Add the mass g of hydrochloric acid

The mass g of the remaining solid

Through the third set of data and the fourth set of data, we know that HCL has 45 50g, and the remaining solids are the same, indicating that the solids have been completely reacted in 45ghCl, hydrochloric acid is excessive or just completely reacted, and CaCO3 is completely reacted, and the total reaction is 10-1g=9g, that is, the mass fraction is 9 10=90%.

Through two sets of data, we know that for every 10g of hydrochloric acid, 2g of solids are reacted, so 45g is added, and 9g of solids are just reacted, so there is no hydrochloric acid left!

Then: sufficient hydrochloric acid, 9g of calcium carbonate in 10g of solid is completely reacted.

2hcl+caco3==cacl2+h2o+co2

100g 44g

9 g xgx=

The conditions of this question may not be complete, and you need to know what the mass of "aluminum and magnesium of equal mass" is.

Suppose that the masses of "aluminum and magnesium of equal mass" are m, and write the chemical equation for the reaction of magnesium, aluminum, and hydrochloric acid to produce hydrogen, respectively.

The equation for magnesium is that magnesium produces one hydrogen, that is, magnesium added to 24 produces hydrogen with 2, and the actual mass added is (24-2) 24=11 12, which is 11m 12.

The equation of aluminum is that two aluminums produce three hydrogen, that is, 27*2=54 aluminum is added to produce 3*2=6 hydrogen, and the actual increase in mass is (54-6) 54=8 9, that is, 8m 9.

8m 9<11m 12, so you need to add aluminum on the aluminum side. Let the total mass of aluminium be m', then in order to balance the balance twice, 8m is required'/9=11m/12，m'=33m/32。However, aluminum with a mass of M has already been added, so a new mass of M needs to be added'-m=m32 of aluminum.

The final answer is to add aluminium of mass m 32 to one side of the aluminium, where m is the mass of the balance that has been added twice at the beginning.

Note: Since the goal of the question is to clarify a sufficient amount of dilute hydrochloric acid, it is assumed that aluminum and magnesium will react completely after being added

Although the mass of aluminum-magnesium added is equal, the amount of substance is not equal, so the quality of the hydrogen released after a complete reaction is not the same. So how much worse?

Let the masses of magnesium and aluminum be a, then a 24 the amount of magnesium that emits hydrogen.

a 27 The amount of aluminium is a 27*

Then the difference between the ends of the balance is a 18-a 24 amount of hydrogen. i.e. a 72 grams of aluminum into 27 grams of hydrogen is released, so a 8 grams of aluminum needs to be put in.

Add another 18 grams of magnesium to the right plate:

3mg+6hcl→3mgcl2+3h2

2al+6hcl→2alcl3+3h2

Equal hydrogen gas is generated.

The amount of magnesium required is: 3 24 = 72

The amount of aluminum required is: 2 27 = 54

Therefore, the amount of magnesium that needs to be added to the right disc is: 72-54 = 18g Basis: equal mass, and only hydrogen runs out in the end.

There are so many uncertainties. Is it completely reactive or is there an excess of metal? Otherwise, there is no way to do it.

In this world, it is impossible to find a specific quality.

Too much trouble to count.

I hate analytical chemistry the most ==

It is essential to know the excess or small amount of magnesium aluminum.

Fewer conditions. I don't know the quality, I can't count.

It can be seen that the two solutions react completely. First of all, it is necessary to determine the conservation of the K element, 10%x56g= , which is , so KCl is also mol, and the mass obtained is mol x( 39+ g. The mass of the solution in the container is 56+=.

The mass fraction of the solute in the solution is 10%.