What is the theoretical yield of by the reaction 3Ca(OH)2(s)+2H3PO4(l)→Ca3(PO4)2(s)+3H2O(l)(4A.4)(4A.4)3Ca(OH)2(s)+2H3PO4(l)→Ca3(PO4)2(s)+3H2O(l) when 10.00 g Ca(OH)2 and 10.00 g H3PO4 are mixed? [f.w. Ca(OH)2 = 74.10 u; m.w. H3PO4 = 97.99 u; f.w. Ca3(PO4)2 = 310.18 u]
Question
What is the theoretical yield of by the reaction 3Ca(OH)2(s)+2H3PO4(l)→Ca3(PO4)2(s)+3H2O(l)(4A.4)(4A.4)3Ca(OH)2(s)+2H3PO4(l)→Ca3(PO4)2(s)+3H2O(l) when 10.00 g Ca(OH)2 and 10.00 g H3PO4 are mixed? [f.w. Ca(OH)2 = 74.10 u; m.w. H3PO4 = 97.99 u; f.w. Ca3(PO4)2 = 310.18 u]
Solution
To calculate the theoretical yield, we first need to determine the limiting reactant.
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Convert the mass of each reactant to moles:
- For Ca(OH)2: 10.00 g / 74.10 g/mol = 0.135 mol
- For H3PO4: 10.00 g / 97.99 g/mol = 0.102 mol
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Determine the mole ratio from the balanced equation:
- The balanced equation tells us that 3 moles of Ca(OH)2 react with 2 moles of H3PO4.
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Compare the mole ratio of the reactants with the mole ratio of the balanced equation:
- The ratio of Ca(OH)2 to H3PO4 in our reaction is 0.135 mol / 0.102 mol = 1.32
- The ratio in the balanced equation is 3/2 = 1.5
Since the ratio of our reactants is less than the ratio in the balanced equation, Ca(OH)2 is the limiting reactant.
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Use the stoichiometry of the reaction to find the moles of product formed:
- The balanced equation tells us that 3 moles of Ca(OH)2 produce 1 mole of Ca3(PO4)2.
- So, 0.135 mol of Ca(OH)2 will produce 0.135 mol / 3 = 0.045 mol of Ca3(PO4)2.
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Convert the moles of product to grams:
- 0.045 mol * 310.18 g/mol = 13.96 g
So, the theoretical yield of Ca3(PO4)2 is 13.96 g.
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