Question

1. Write the balanced chemical reaction for silver metal reacting with phosphoric acid to form

silver phosphate and hydrogen gas. What is the percent yield if 48.99 grams of silver
phosphate are produced from the reaction of 50.00 grams of silver with excess phosphoric
acid?
Balanced Equation:

Answer 1

75.51%

Solution and Explanation:

Balanced equation: 6Ag(s) + 2H₃PO₄(aq) → 2Ag₃PO₄(aq) + 3H₂(g)

To determine the percent yield of silver phosphate we will use the following steps;

Step 1: Determine the number of moles of Silver metal used

Mass of silver metal = 50.00 grams

Molar mass of silver is 107.87 g/mol

Therefore;

Moles of silver = 50.0 g ÷ 107.87 g/mol

= 0.464 moles

Step 2: Determine moles of silver phosphate produced

From the equation, 6 moles of Ag reacts with 2 moles of phosphoric acid to yield 2 moles of silver phosphate.

Thus, the mole ratio of silver metal to silver phosphate is 6 : 2

Therefore;

Moles of Ag₃PO₄ = Moles of Ag × 2/6

= 0.464 moles × 2/6

= 0.155 moles

Step 3: Calculate the theoretical mass of silver phosphate produced

Mass = Number of moles × Molar mass

Number of moles of Ag₃PO₄= 0.155 moles

Molar mass of Ag₃PO₄ =418.58 g/mol

Therefore;

Theoretical mass of Ag₃PO₄ = 0.155 moles × 418.58 g/mol

= 64.88 g

Step 4: Calculate the percent yield of Ag₃PO₄

`percent yield=(Actual mass)/(Theoretical mass)(100)`

Actual mass of Ag₃PO₄ = 48.99 g

Theoretical mass of Ag₃PO₄ = 64.88 g

Therefore;

% yield of Ag₃PO₄ = (48.99 g ÷ 64.88 g)×100%

= 75.51%

Hence, the percent yield of Ag₃PO₄ is 75.51%