If 0.500 mol of CaCl₂is mixed with 0.200 mol Na₃PO₄, the maximum amount in moles of Ca₃(PO₄)₂ that can be formed is:

Question

Quizplus · Accepted Answer

The balanced reaction is 3CaCl₂ + 2Na₃PO₄ → Ca₃(PO₄)₂ + 6NaCl. From the stoichiometry, 3 moles of CaCl₂ react with 2 moles of Na₃PO₄ to form 1 mole of Ca₃(PO₄)₂. Therefore, 0.500 mol CaCl₂ would require 0.333 mol Na₃PO₄, but only 0.200 mol Na₃PO₄ is available, making Na₃PO₄ the limiting reactant. Thus, 0.200 mol Na₃PO₄ can produce 0.100 mol Ca₃(PO₄)₂.

If 0500 Mol of CaCl2 is Mixed with 0

If 0500 Mol of CaCl₂is Mixed with 0