# How to Solve Simple Stoichiometry Problems Involving Solvents?

• AmazingLight
In summary: I can see how to do the question.In summary, to solve the question, you need to know the concentration of the NaCl solution, the mass of silver chloride precipitated, and the mole fraction of silver nitrate in the original solution.
AmazingLight
Hey guys,

I have two stoichiometry questions to do with solvents. Both of them are very simple and I am unable to complete them at the moment because of some silly error in my method, I am sure. Can you please explain the method and how to do the question in some detail? I really appreciate your help.6) What volume of 0.250M Sulphuric Acid (H2SO4) is needed to react completely with 13.5g of Sodium Hydroxide(NaOH)?

According to this equation: 2NaOH(aq) + H2SO4 = Na2SO4(aq) + 2H2O(l) 7) 200mL of 0.105 M silver nitrate is added drop wise to 25mL of sodium chloride until all the chloride precipitates as silver chloride:

AgNO3(aq) + NaCl(aq) = AgCl(s) + NaNO3(aq)

* What is the concentration of the original sodium chloride solution?

* What is the mass of silver chloride precipitated?

Here is a photo to clear any confusion (if any). If you could please explain question [8] too, it'll be great. Thanks.

In order to receive help, you must show some work in solving these problems.

SteamKing said:
In order to receive help, you must show some work in solving these problems.

Ok...so what's wrong with what I've done.

Concentration of NaCl is OK, but I have no idea what you did later and why. Volume of the NaCl solution was given, so there is no need to calculate it, and the question asked to calculate something else.

Would this be right then?

6) 2NaOH(aq) + H2SO4 = Na2SO4(aq) + 2H2O(l)
n(NaOH) = m/M = 13.5/40 = 0.3375mol
n(H2SO4) = n(NaOH) * 0.5 = 0.16875 mol
n(H2SO4) = x L * 0.250molL^-1
0.16875/0.250 = x = 0.675L

0.675L of sulfuric acid is needed.

7) AgNO3(aq) + NaCl(aq) = AgCl(s) + NaNO3(aq)
n(AgNO3) = 0.105/5 = 0.021 mol
n(NaCl) = n(AgNO3) = 0.021 mol
n(NaCl) = 0.025L * y molL-1
y = 0.021mol/ 0.025L = 0.84

Concentration of NaCl solution = 0.84M

n(AgCl) = n(NaCl) = 0.021mol
m = nM = 0.021 * (107.8+35.5) = 3.01g

Mass of precipitate = 3.01g

OK now.

## What is stoichiometry?

Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction, based on the law of conservation of mass.

## How do you calculate the amount of product formed in a reaction?

To calculate the amount of product formed, you need to know the balanced chemical equation and the amounts of the reactants. Using stoichiometry, you can convert the given amount of reactant into the corresponding amount of product.

## What is the mole ratio in stoichiometry?

The mole ratio is the ratio of the coefficients of the reactants and products in a balanced chemical equation. It is used to convert between the amounts of different substances involved in a reaction.

## What is the difference between theoretical yield and actual yield?

Theoretical yield is the calculated amount of product that should be formed in a reaction, based on stoichiometric calculations. Actual yield is the amount of product that is actually obtained from the reaction in a laboratory setting. The actual yield is often less than the theoretical yield due to factors such as incomplete reactions or loss of product during the experiment.

## How do you determine the limiting reactant in a reaction?

The limiting reactant is the reactant that is completely consumed in a chemical reaction. To determine the limiting reactant, you need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces the least amount of product is the limiting reactant.

Replies
1
Views
6K
Replies
3
Views
6K
Replies
8
Views
4K
Replies
4
Views
3K
Replies
2
Views
4K
Replies
8
Views
7K
Replies
3
Views
10K
Replies
4
Views
22K
Replies
2
Views
4K