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1] How many grams of calcium phosphate can be produced from the reaction of 2.50 l of 0.250 m Calcium chloride with and excess of phosphoric acid? calcium chloride + phosphoric acid --> calcium phosphate + hydrochloric acid answer 2] How many milliliters of 1.50 m Nitric acid is required to react with 100.0 g of cuprous oxide 14 HNO3 + 3 Cu2O 6 Cu(NO3)2 + 2 NO+ 7 H2O Answer
Accepted Answer
1] Balanced Equation:
3CaCl₂(aq) + 2H₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6HCl(aq)
Calculations:
Moles of CaCl₂: (2.50 L)(0.250 mol/L) = 0.625 mol CaCl₂
Moles of Ca₃(PO₄)₂: (0.625 mol CaCl₂)(1 mol Ca₃(PO₄)₂ / 3 mol CaCl₂) = 0.208 mol Ca₃(PO₄)₂
Grams of Ca₃(PO₄)₂: (0.208 mol Ca₃(PO₄)₂)(310.18 g/mol) = 64.5 g Ca₃(PO₄)₂
2] Balanced Equation:
14 HNO₃(aq) + 3 Cu₂O(s) → 6 Cu(NO₃)₂(aq) + 2 NO(g) + 7 H₂O(l)
Calculations:
Moles of Cu₂O: (100.0 g Cu₂O)(1 mol Cu₂O / 143.09 g/mol) = 0.699 mol Cu₂O
Moles of HNO₃: (0.699 mol Cu₂O)(14 mol HNO₃ / 3 mol Cu₂O) = 3.28 mol HNO₃
Volume of HNO₃: (3.28 mol HNO₃)(1 L / 1.50 mol) = 2.19 L HNO₃
Milliliters of HNO₃: (2.19 L)(1000 mL/L) = 2190 mL HNO₃
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