Answer to Question #105176 in General Chemistry for maria zal

Question

what mass (in grams) of strontium phosphate is formed from the reaction of 29.0 mL of 0.0520 M SrCl2 solution with 12.0 mL of 0.110 M Na3PO4 Solution?

Answer 1

What mass (in grams) of strontium phosphate is formed from the reaction of 29.0 mL of 0.0520 M SrCl₂ solution with 12.0 mL of 0.110 M Na₃PO₄ Solution?

Solution:

The equation of the chemical reaction is following:

3SrCl₂ + 2Na₃PO₄ = Sr₃(PO₄)₂ + 6NaCl

We can calculate the mass of SrCl₂ using given values:

m(SrCl₂) = Cm(SrCl₂) * M(SrCl₂)*V = 0.0520 mol/L * (87.62+2*35.5) g/mol*0.029 L = 0.2392 (g)

The mass of Na₃PO₄ equals:

m(Na₃PO₄) = Cm(Na₃PO₄) * M(Na₃PO₄)*V = 0.110 mol/L * (3*23+1*31+4*16) g/mol*0.012 L = 0.2165 (g)

We can determine which of these reagents given in excess according to chemical equation:

0.2392 g X g

3SrCl₂ + 2Na₃PO₄ = Sr₃(PO₄)₂ + 6NaCl

3*158.62 2*164

Make a proportion:

0.2392/475.86 = X/328

where:

X = m(Na₃PO₄) = 0.2392*328/475.86 = 0.1649 (g)

Thus, Na₃PO₄ was given in excess and SrCl₂ was given in lack.

So we can calculate the mass of product Sr₃(PO₄)₂ using the mass of SrCl₂:

0.2392 g X g

3SrCl₂ + 2Na₃PO₄ = Sr₃(PO₄)₂ + 6NaCl

3*158.62 452.86

Make a proportion:

0.2392/475.86 = X/452.86

where:

X = m(Sr₃(PO₄)₂) = 0.2392*452.86/475.86 = 0.2276 (g)

Answer: m(Sr₃(PO₄)₂) = 0.2276 g.