Answer:
Explanation:
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In this case, according to the given question about stoichiometry, it is possible for us to calculate the required moles of water that will be produced by 12 grams of hydrogen, by using the molar mass of this reactant (2.02 g/mol as it is diatomic) and the 2:2 mole ratio in the chemical equation by solving the following setup:
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75.51%
<h3>Solution and Explanation:</h3>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
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%