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 %

Step-by-step explanation:

The balanced equation for the reaction between silver metal and phosphoric acid is given by;

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

We are required to calculate the percentage yield.

First we calculate the actual yield when 50.0 g reacted with acids;

Step 1: Moles of silver that reacted

Moles = mass ÷ molar mass

Molar mass of Ag = 107.868 g/mol

Thus;

Moles of Ag = 50.0 g ÷ 107.868 g/mol

= 0.464 moles

Step 2: Moles of Silver phosphate produced

From the equation 6 moles of silver reacts to produce 2 moles of Silver phosphate.

Therefore, 0.464 moles of Ag will produce;

= 0.464 moles × 2/6

= 0.155 moles

Step 3: Theoretical mass of silver phosphate produced by 50.0 g of silver

Mass = moles × Molar mass

Molar mass = 418.58 g/mol

Mass of silver phosphate = 0.155 moles × 418.58 g/mol

= 64.88 g

Step 4: Percentage yield of silver phosphate

% yield = (Actual yield ÷Theoretical yield) × 100%

= (48.99 g ÷ 64.88 g)×100%

= 75.51 %

Thus, the percentage yield of silver phosphate is 75.51%