Please help with chemistry question

Question

Calcium chloride (aq) reacts with sodium carbonate (aq) to from solid calcium carbonate and aqueous sodium chloride. Determine the volume of a 2.00 M Calcium chloride solution would be needed to exactly react with 0.0650 L of 1.50 M Na₂CO₃.

Answer 1

Write the balanced equation:

CaCl₂(aq) + Na₂CO₃(aq) ==> CaCO₃(s) + 2NaCl(aq) ... balanced equation

To find the volume of CaCl2 solution, we must first find the number of moles of CaCl₂. And to find moles of CaCl₂, we first need to know the moles of Na₂CO₃ that are present.

moles Na₂CO₃ = 0.0650 L x 1.50 mol/L = 0.0975 moles Na₂CO₃

moles CaCl₂ needed = 0.0975 moles Na₂CO₃ x 1 mol CaCl₂/mole Na₂CO₃ = 0.0975 mol CaCl₂

Volume of CaCl₂ needed: (x L)(2.00 mol/L) = 0.0975 moles

x = 0.04875 L = 48.8 mls (3 sig. figs.)