Question

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

Answer 1

Answer:

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

$percent yield=\frac{Actual mass}{Theoretical mass}(100)$

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%