Two chemical equation calculation problems for junior high school students

1. 170g has been decomposed with a mass fraction of 14%.

2, if memory serves, it is 40. The mass fraction is approximately equal to 72%.

Let the original mixture Kclo3 be Xg, and generate O2 according to the law of mass repentance and Hengheng, and the mass is 187+20-169=38g

2kclo3===mno2 plus medium excitation ===2kCl+3O2x 38 solution: x=97

Then the mixture KCl is 187-97=90

The mass of K in KCLO3 is 97*39 245*100%=The mass of K in KCL is 90*39

So the fraction of the mass of k is (

According to the title, the quality of mno2 is because only he is insoluble in water, and then kclo3 and kcl are the quality of the socks.

Then it can be seen that the mass of the k-element of their mixture is.

Let kclo3 in the mixture be x, then we can know x*39 245+ (to find x, it is enough to find the amount of O2 according to the chemical equation.

Hope it helps.

1. Let the precipitated crystal mass be x, then it includes 5g of anhydrous copper sulfate powder and (x-5)g original saturated solution, so the mass of copper sulfate = 5 + (x-5) * 25 125 = 4 + x 5 g, that is, the mass of copper sulfate in the crystal = x * 160 250 = 4 + x 5, and the solution is x =

2. The original solution should be a saturated solution, right? where solute mass = 500 * 55 155 = and water mass =

Let the precipitated copper sulfate crystal mass be x, then the remaining copper sulfate mass in the solution = and the remaining water mass = , this solution is 30 saturated solution, so (, solution.

x=175g。

1.Solution: The mass of HCl in 40ml of hydrochloric acid is:

Let the mass of sodium hydroxide in the 10 grams sample be X, and the mass of sodium chloride generated after the reaction is YNOH+2HCl===NaCl+H2O

x y40：73=x： x=

73：：y y=

The mass of sodium chloride in 10 g of sample is:

The mass fraction of sodium chloride in the sample was: x100%=

The mass of sodium chloride in the solution obtained after the reaction is:

The mass of the solution obtained after the reaction is 10g+20g+

The mass fraction of solute in the solution after the reaction is: x100%=A: omitted. 2。Solution: Let the mass of the solute in the sodium carbonate solution be x and the mass of Ca(OH)2 in the original mixture be Y

na2co3+ca（oh）2===2naoh+caco3x y 12g-（

74：80=y：（ y=

106：74=x： x=

The mass of CaCO3 formed after the reaction is:

The mass of the solution obtained after the reaction is:

The solute mass fraction of the filtered solution is: 12g 120g x100% = 10% Answer: omitted.

The following answer is very good!!

(1) HCl FeCl3 (At the beginning, hydrochloric acid consumes sodium hydroxide and does not produce precipitation. As sodium hydroxide increases, hydrochloric acid is depleted and begins to react with ferric chloride to form iron hydroxide precipitates).

2) According to the data analysis in the table: between 40-60g to produce precipitation between 60-80g to produce precipitation; Then between 40-60, sodium hydroxide has been reacting with ferric chloride, that is, at 40g, hydrochloric acid is just consumed At the same time, it is known that every 20g of sodium hydroxide consumed, a precipitate is generated, so that 160g should be generated precipitate, and the question should be changed to that is, at exactly 160, ferric chloride is consumed.

HCl consumes 40GNOH solution FeCl3 consumes 120g of NAOH solution

Mass of HCL = 40 5% 40

Mass of FeCl3 = 120 5% 40 3 Write the chemical equation and the result can be obtained according to the ratio)

According to your hair**, the residue should contain excess dilute hydrochloric acid and ferric chloride, both of which are brownish-yellow.

Because, just added sodium hydroxide, no precipitation occurred, indicating that ferric chloride has not reacted with sodium hydroxide, but the excess dilute hydrochloric acid reacts with sodium hydroxide first to generate sodium chloride and water, iron chloride, hydrochloric acid, sodium hydroxide are easily soluble in water, naturally there is no precipitation.

As the sodium hydroxide is added, the excess dilute hydrochloric acid in the residual solution is consumed, and the ferric chloride in the residual solution begins to react with the sodium hydroxide to form a brown precipitate iron hydroxide.

Reaction Equation:

fe2o3+6hcl====2fecl3+3h2ohcl+naoh====nacl+h2o

fecl3+3naoh====3nacl+fe(oh)3↓

The solute is FeCl3, ferric chloride, and the solution of ferric chloride is yellow. The second question can be seen from the figure that 160g of sodium hydroxide can no longer generate more iron hydroxide precipitation, indicating that iron hydroxide is the most precipitate that can be generated, so it can be concluded by comparing the molecular mass equation that the iron hydroxide is generated from iron chloride. And so much ferric chloride is the solute in the original 100g solution.

So I think ferric chloride is.

Because, just added sodium hydroxide, there is no precipitation, indicating that the excess dilute hydrochloric acid reacts with sodium hydroxide first to generate sodium chloride and water, and when the excess dilute hydrochloric acid in the residual solution is consumed, ferric chloride begins to react with sodium hydroxide to generate reddish-brown precipitate iron hydroxide.

So, (1) HCl, FeCl3

2.Let the mass of FeCl3 be X Naoh=(60g-40g)*5%=1g

fecl3+3naoh=3nacl+fe(oh)3↓120

x 1gx is approximately equal to.

I believe you understand the basic calculation format.

Methane is CH4

Solution:1 12g ÷(12×2）／（12×2+6×1+16）×100%=23g

Answer: - 2Assuming that at least the mass of oxygen required is xg, then CH4 + 2 O2 CO2 + 2 H2O16 2 32

x∴ ＝x ／2×32

x Answer: ——

n (alcohol) =

m (alcohol) =

Methane) + 2O2 = = = > CO2 + 2H20

Solve using coefficient ratios.

1. Let the alcohol be mg, and the column formula is 24 46*m=12, and calculate m=23g2, methane should be ch4, ch4+2o2=co2+2h2o16 32

1.First, find the mass fraction of carbon in alcohol, divide 12 grams by the mass fraction.

Molecular methane and 2 molecules of oxygen are fully combusted, with a molecular weight of 16 for methane and 32 for oxygenSo the methane is fully burned and oxygen is needed. Methane; The chemical formula is CH4

1.Calculations about the relative molecular mass. First, find the mass fraction of carbon in alcohol according to the chemical formula, and then calculate how many grams of alcohol contain 12 grams of carbon.

2.Calculations using chemical equations. (1) Set the unknown quantity (2) Write the chemical equation (3) Write the relative molecular mass of the related substance and the ratio of the known quantity and unknown quantity (4) Solve and solve (5) Write the answer concisely.

Calculate the relative atomic mass of the element carbon and divide it by twelve

1.The alcohol mass fraction is 46 and the carbon is 12The calculation process is 12 (12 46) = 46 grams.

The required alcohol is 46 grams. , the methane mass fraction is 16, the 2O2 mass fraction is 64, and the oxygen mass is x16 Solution x = oxygen required.

Relative molecular weight 46

So 23g

2 ch1??

1、fe3o4——4co2

xx=232 / 32

Mass fraction = (x 10) 100% = Fe3O4 – 4Fe

xx=232 × / 224

Mass fraction = (x 16) 100% =

1.Solution: Let the mass of zinc be x

zn+2hcl=zncl+h2

x solution x = answer: . .

2.Solution C+O2=CO2 CH4+2O2=CO2+H2O

m m (c combustion generated) m m (ch4) combustion generated).

CO2 from m(c) combustion = 44m 12= CO2 from combustion of m(CH4) = 44m 16=

CO2 from m(c) combustion m(ch4) CO2).

Answer:..