Identify the coefficient of Ca(OH)₂(aq)when the following reaction is balanced,and determine the number of millilitres of 0.00300 mol L-1 phosphoric acid required to neutralize 45.00 mL of 0.00150 mol L^-1 calcium hydroxide. (unbalanced)Ca(OH)₂(aq)+ H₃PO₄(aq)→ Ca₃(PO₄)₂(aq)+ H₂O(l) A)Coefficient 1,volume 3.04 mL B)Coefficient 3,volume 15.0 mL C)Coefficient 2,volume 22.5 mL D)Coefficient 4,volume 33.8 mL

Question

Quizplus · Accepted Answer

The coefficient of Ca(OH)2 is 1, and the coefficient of H3PO4 is 3 in the balanced equation: 3Ca(OH)2(aq) + 2H3PO4(aq) → Ca3(PO4)2(s) + 6H2O(l). Therefore, one mole of Ca(OH)2 will react with three moles of H3PO4. The number of moles of H3PO4 in 45.00 mL of 0.00150 mol L-1 solution is 0.0675 mmol. To neutralize this, we need 0.0675/3 = 0.0225 mmol of H3PO4, which is present in 0.0225/0.00300 = 7.50 mL of 0.00300 mol L-1 H3PO4. Therefore, the best choice is B, which is closest to this value.

Identify the Coefficient of Ca(OH)2(aq)when the Following Reaction Is Balanced,and

Identify the Coefficient of Ca(OH)₂(aq)when the Following Reaction Is Balanced,and