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

36 grams of tin (IV) phosphate Sn3(PO4)4 is mixed with an excess of sodium carbonate Na2CO3. In the lab 29.8 grams of tin (IV) carbonate Sn(CO3)2 are actually formed. What is the percent yield?

Question

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

36 grams of tin (IV) phosphate Sn3(PO4)4 is mixed with an excess of sodium carbonate Na2CO3. In the lab 29.8 grams of tin (IV) carbonate Sn(CO3)2 are actually formed. What is the percent yield?

BlackTom AI · Accepted Answer

We start by identifying the stoichiometry of the given balanced equation: Sn3(PO4)4 + 6 Na2CO3 → 3 Sn(CO3)2 + 4 Na3PO4. This tells us that 1 mole of tin(IV) phosphate produces 3 moles of tin(IV) carbonate.

Next, we calculate the molar masses needed for converting between grams and moles:
- Sn3(PO4)4: 3 Sn atoms (3 × 118.71 g/mol) plus 4 PO4 groups (4 × [P 30.97 g/mol + 4 × O 16.00 g/mol]) = 356.13 g/mol + 379.88 g/mol = 736.01 g/mol.
- Sn(CO3)2 (tin(IV) carbonate): Sn 118.71 g/mol plus 2 CO3 groups (2 × [C 12.01 g/mol + 3 × O 16.00 g/mol] = 2 × 60.01 g/mol) = 118.71 g/mol + 120.02 g/mol = 238.73 g/mol.

Now determine the theoretical yield from the given starting mass of Sn3(PO4)4 (36.0 g):
- Moles of Sn3(PO4)4 = 36.0 g ÷ 736.01 g/mol ≈ 0.0489 mol.
- From the 1:3 stoichiometry, moles of Sn(CO3)2 formed theoretically = 3 × 0.0489 ≈ 0.1467 mol.
- Convert to grams: theoretical mass of Sn(CO3)2 = 0.1467 mol × 238.73 g/mol ≈ 35.0 g (more precisely about 35.0–35.04 g depending on rounding).

The percent yield is defined as (actual yield ÷ theoretical yield) × 100%. Here the actual yield is 29.8 g.
- Percent yield = (29.8 g ÷ 35.0 g) × 100% ≈ 0.851 × 100% ≈ 85%.

Key checks and considerations:
- Ensure the correct stoichiometric coefficients are used when converting moles of the reactant to moles of the product.
- Use accurate molar masses for Sn3(PO4)4 and Sn(CO3)2 to minimize rounding errors in the theoretical yield.
- The reaction mixture having excess Na2CO3 means the limiting reagent is Sn3(PO4)4, so the theoretical yield hinges on its amount.
- Small variations in rounding during intermediate steps can slightly alter the final percent yield, but they should still be around the 85% range given these inputs.

This workflow yields a percentage close to the reported value, illustrating how to connect mass input to theoretical product mass and then to percent yield.

Chem Earth B LT 6.4 Content Check

Sn3(PO4)4 + 6 Na2CO3 → 3 Sn(CO3)2 + 4 Na3PO4 36 grams of tin (IV) phosphate Sn3(PO4)4 is mixed with an excess of sodium carbonate Na2CO3. In the lab 29.8 grams of tin (IV) carbonate Sn(CO3)2 are actually formed. What is the percent yield?

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