Question

Na₃PO₄(aq) + Ca(NO₃)₂(aq) → NaNO₃(aq) + Ca₃(PO₄)₂(s)

Determine the amount of NaNO₃(aq) formed in the reaction if 3.30 moles of Ca(NO₃)₂(aq) reacts with an excess of Na₃PO₄(aq) and the percent yield of NaNO₃(aq) is 71.0%.

A. 6.24 moles

B. 9.30 moles

C. 4.69 moles

D. 7.50 moles

E. 8.96 moles

Answer 1

Balanced equation for this reaction is,

2Na3PO4 + 3Ca(NO3)2 -----> Ca3(PO4)2 + 6NaNO3

3 moles of Ca(NO3)2 gives 6 moles of NaNO3.

1 mole of Ca(NO3)2 gives 2 moles of NaNO3.

3.30 moles of Ca(NO3)2 will give 2×3.30 moles NaNO3.

= 6.60 moles NaNO3.

Actual yield ° percent yield/100 × theoretical yield.

Actual yield = 71/100 × 6.60

= 4.69 moles