chemistry question

Question

Consider the reaction 3CaCl₂(aq) + 2Na₃PO₄(aq) --> Ca₃(PO₄)₂(s) + 6 NaCl(aq) (molar mass of calcium phosphate is 310.18 g/mol)

What volume of 0.500 M Na₃PO₄ solution is needed to react with 275 mL of 0.150 M CaCl₂? _____ mL (Give answer to correct sig figs in units of mL)

Answer 1

3CaCl₂(aq) + 2Na₃PO₄(aq) ==> Ca₃(PO₄)₂(s) + 6NaCl(aq) ... balanced equation

From the balanced eq. we see that it takes 2 moles Na₃PO₄ to react with 3 moles CaCl₂. So, from the volume and molarity of CaCl₂ given, we can find the moles of Na₃PO₄ needed.

moles CaCl₂ present = 275 ml x 1 L/1000 ml x 0.150 mol/L = 0.04125 moles CaCl₂

moles Na₃PO₄ needed = 0.04125 mol CaCl₂ x 2 mol Na₃PO₄ / 3 mol CaCl₂ = 0.0275 moles Na₃PO₄

Volume Na₃PO₄ needed:

(0.500 mol/L)(x L) = 0.0275 moles

x = 0.0550 L = 55.0 ml (to 3 significant figures)