If 6.70 g of zinc is reacted with 27.5 g of silver nitrate, the products will be zinc nitrate and silver metal. What is the theoretical mass of silver that will be produced? a. Write a balanced equation

b. Interpret the equation by
particles and by moles

c. What is the limiting reactant?

Given:

Molar mass AgNO3 =

Needed:

unknown:

available:

d. What is the theoretical mass of silver that will be produced?

e. If the actual yield of silver was 14.1 g, what is the percent yield of the
reaction?

Explanation:

a. Write a balanced equation

The balanced equation for the reaction is given as;

Zn + AgNO3 → Ag + ZnNO3  

b. Interpret the equation by  particles and by moles

In this equation, 1mol of Zn reacts with 1 mol of AgNO3 to from 1 mol of Ag and 1 mol of ZnNO3

1 mol = 6.022 * 10^23 particles

c. What is the limiting reactant?

The mol ratio between both reactants is 1 : 1 from the equation.

Converting the masses given to ml;

Number of mol = Mass / Molar mass

Zn;

Number of mol = 6.70 / 65.38 = 0.1025 mol

AgNO3;

Number of mol = 27.5 / 169.87 = 0.1619 mol

The limiting reactant in this reaction is Zn. This is because Zn would get used up before AgNO3 according to the number of moles calculated.

d. What is the theoretical mass of silver that will be produced?

1 mol of Zn produces 1 mol of Ag

0.1025 mol of Zn would also produce 0.1025 mol mol of Ag

Mass = Molar mass * Number of Moles = 107.87  * 0.1025

Mass = 11.05 g

e. If the actual yield of silver was 14.1 g, what is the percent yield of the

reaction?

Percent yield = Actual Yield / Theoretical yield * 100%

Percent yield = 11.05 / 14.1  * 100 = 78.36%

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