Question

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

Answer 1

Answer:

75.51 %

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%