Calculate AgCl, NaNO3 from 1.3g NaCl + 3.5g AgNO3

In summary, the conversation discusses a chemical reaction involving NaCl and AgNO3, and asks for the resulting amounts of AgCl and NaNO3, as well as the limiting reactant and any excess reactant. The calculations show that AgNO3 is the limiting reactant and will yield 3.0g of AgCl and 1.8g of NaNO3. However, the answer is marked incorrect on Mastering Chemistry, with the correct mass of NaCl being 9.2*10^-2g. A stoichiometry calculator is attached with the calculated results.
  • #1
Faiien
11
0

Homework Statement


( 1.3 g NaCl )+( 3.5 g AgNO3) ---> (xg AgCl) + (yg NaNO3)

How many grams of each product result from the following reactions, and how many grams of which reactant is left over?

Enter your answers numerically separated by commas. Express your answer using two significant figures.
mNACl, mAgCl, mNaNO3 = g

Homework Equations





The Attempt at a Solution


1.3g NaCl*(1mole of NaCl/58.44g)=0.022 AgCl
3.5g AgNO3*(1mole of AgNO3/169.87g)=0.0206(limiting reagent)
0.0206 mols AgCl (143.32g/1 mol of AgCl)=3.0g
0.0206 mols NaNO3 (84.99/1 mol of NaNO3)=1.8g

I've figured out that AgNO3 is the limiting reactant which will yield 0.0206 moles of AgCl and NaNO3. I've calculated that 0.0206 moles of AgNO3 will yield 3.0g of AgCl and 1.8g of NaNO3.
This seems to be correct because 1.3g+3.5g = 3.0g+1.8g.
When I enter 1.3,3.0,1.8 into masteringchemistry, it tells me I'm incorrect and I don't understand why. I would really appreciate some help.


Doh! Solved it. Sorry everyone.
 
Last edited:
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  • #2
Dude I honestly made this account just to help you...
first off Mastering chem is a pain in the arse, further more, when you answered
the question you answered correctly, except for the Mass of NaCl
which is actually 9.2*10^-2 for some reason...
 
  • #3
Theodocious said:
Dude I honestly made this account just to help you...

Year and half after the question was asked. Sigh.

Attached are results as calculated by EBAS stoichiometry calculator.
 

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1. How do you calculate the amount of AgCl and NaNO3 produced from a given amount of NaCl and AgNO3?

To calculate the amount of AgCl and NaNO3 produced, we first need to determine the molar ratio between the reactants and the products. In this case, the molar ratio is 1:1 between NaCl and AgCl, and 1:1 between AgNO3 and NaNO3. Next, we calculate the moles of each reactant by dividing their given amounts by their respective molar masses. Then, using the molar ratio, we can determine the moles of AgCl and NaNO3 produced. Finally, we convert the moles to grams by multiplying by their respective molar masses.

2. What is the chemical equation for the reaction between NaCl and AgNO3?

The chemical equation for the reaction between NaCl and AgNO3 is: NaCl + AgNO3 → AgCl + NaNO3

3. How do you find the molar mass of NaCl and AgNO3?

The molar mass of a compound is calculated by adding the molar masses of each element in the compound. The molar mass of NaCl is 58.44 g/mol (22.99 g/mol for Na + 35.45 g/mol for Cl) and the molar mass of AgNO3 is 169.87 g/mol (107.87 g/mol for Ag + 14.01 g/mol for N + 3(16.00 g/mol) for O).

4. What is the theoretical yield of AgCl and NaNO3 from 1.3g NaCl + 3.5g AgNO3?

The theoretical yield is the maximum amount of product that can be produced from a given amount of reactants. In this case, the theoretical yield of AgCl and NaNO3 can be calculated by using the molar ratio and converting the moles to grams. The theoretical yield of AgCl is 1.3g and the theoretical yield of NaNO3 is 3.5g.

5. What factors can affect the actual yield of AgCl and NaNO3 from this reaction?

The actual yield may differ from the theoretical yield due to various factors such as impurities in the reactants, incomplete reactions, and experimental errors. The efficiency of the reaction and the conditions in which it takes place can also affect the actual yield. Additionally, the stoichiometry of the reaction and the presence of any side reactions can also impact the actual yield.

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