Question

Need help with chemistry Atomic Masses: C-12.0 NA-23.0, O-16.0, P-31.0, Sn-119

Answer 1

INFORMATION:

We know that:

- 34.37 g of tin(IV) phosphate is added to 34.05 g of sodium carbonate

- 24.24 g of tin(IV) carbonate are made

And we must find the %yield

STEP BY STEP EXPLANATION:

Balanced equation:

Sn3(PO4)4 + 6Na2CO3 → 3Sn(CO3)2 + 4Na3PO4

1. We must find the amount of tin carbonate produced by tin phosphate

$34.37gSn_3(PO_4)_4\times\frac{1molSn_3(PO_4)_4}{480.03gSn_3(PO_4)_4}\times\frac{3molSn(CO_3)_2}{1molSn_3(PO_4)_4}\times\frac{238.73gSn(CO_3)_2}{1molSn(CO_3)_2}=51.2790gSn(CO_3)_2$

2. We must find the amount of tin carbonate produced by sodium carbonate

$34.05gNa_2CO_3\times\frac{1molNa_2CO_3}{105.9888gNa_2CO_3}\times\frac{3molSn(CO_3)_2}{6molNa_2CO_3}\times\frac{238.73gSn(CO_3)_2}{1molSn(CO_3)_2}=38.3472gSn(CO_3)_2$

3. We must find the theoretical amount of tin carbonate produced

Since sodium carbonate is the limiting reactant, then the theoretical amount of tin carbonate produced would be

$38.3472gSn(CO_3)_2$

4. Finally, the %yield would be

$\text{ \%yield}=\frac{24.24gSn(CO_3)_2}{38.3472gSn(CO_3)_2}\times100=63.2119\text{ \%}$

ANSWER:

1.

$34.37gSn_3(PO_4)_4\frac{1molSn_{3}(PO_{4})_{4}}{480.03gSn_{3}(PO_{4})_{4}}\frac{3molSn(CO_{3})_{2}}{1molSn_{3}(PO_{4})_{4}}\frac{238.73gSn(CO_{3})_{2}}{1molSn(CO_{3})_{2}}=51.2790gSn(CO_3)_2$

2.

$34.05gNa_2CO_3\frac{1molNa_{2}CO_{3}}{105.988,8gNa_{2}CO_{3}}\frac{3molSn(CO_{3})_{2}}{6molNa_{2}CO_{3}}\frac{238.73gSn(CO_{3})_{2}}{1molSn(CO_{3})_{2}}=38.3472gSn(CO_3)_2$

3.

$38.3472gSn(CO_3)_2$

4.

$\operatorname{\%}\text{y}\imaginaryI\text{eld}=\frac{24.24gSn(CO_{3})_{2}}{38.347,2gSn(CO_{3})_{2}}\times100=63.2119\operatorname{\%}$