<h3>
Answer:</h3>
75.51 %
<h3>
Explanation:</h3>
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;
<h3>Step 1: Moles of silver that reacted </h3>
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
<h3>Step 2: Moles of Silver phosphate produced </h3>
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
<h3>Step 3: Theoretical mass of silver phosphate produced by 50.0 g of silver </h3>
Mass = moles × Molar mass
Molar mass = 418.58 g/mol
Mass of silver phosphate = 0.155 moles × 418.58 g/mol
= 64.88 g
<h3>Step 4: Percentage yield of silver phosphate </h3>
% 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%
