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*(1molSn_3(PO_4)_4)/(480.03gSn_3(PO_4)_4)*(3molSn(CO_3)_2)/(1molSn_3(PO_4)_4)*(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*(1molNa_2CO_3)/(105.9888gNa_2CO_3)*(3molSn(CO_3)_2)/(6molNa_2CO_3)*(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}=(24.24gSn(CO_3)_2)/(38.3472gSn(CO_3)_2)*100=63.2119\text{ \%}`

ANSWER:

1.

`34.37gSn_3(PO_4)_4(1molSn_(3)(PO_(4))_(4))/(480.03gSn_(3)(PO_(4))_(4))(3molSn(CO_(3))_(2))/(1molSn_(3)(PO_(4))_(4))(238.73gSn(CO_(3))_(2))/(1molSn(CO_(3))_(2))=51.2790gSn(CO_3)_2`

2.

`34.05gNa_2CO_3(1molNa_(2)CO_(3))/(105.988,8gNa_(2)CO_(3))(3molSn(CO_(3))_(2))/(6molNa_(2)CO_(3))(238.73gSn(CO_(3))_(2))/(1molSn(CO_(3))_(2))=38.3472gSn(CO_3)_2`

3.

`38.3472gSn(CO_3)_2`

4.