If 102 grams of Sn(NO2)4 is reacted with 80.2 grams of Pt3N4, and yields 40.2 grams of Sn3N4. What is the limiting reagent in the reaction and what is the percent theoretical yield?
The correct answer and explanation is :
To solve this problem, we need to determine the limiting reagent and calculate the percent theoretical yield of the reaction. The key steps include identifying the limiting reagent, calculating the theoretical yield, and then finding the percent yield.
Step 1: Write the balanced chemical equation
First, we need to establish the balanced chemical equation for the reaction. Based on the reactants, the reaction can be assumed to proceed as follows:
[
3Sn(NO_2)_4 + 4Pt_3N_4 \rightarrow Sn_3N_4 + 4Pt(NO_2)_4
]
Step 2: Calculate the molar masses of the reactants
Next, calculate the molar masses of the reactants:
- Molar mass of Sn(NO2)4:
[
\text{Sn} = 118.71 \, \text{g/mol}, \, \text{N} = 14.01 \, \text{g/mol}, \, \text{O} = 16.00 \, \text{g/mol}
]
[
\text{Molar mass of Sn(NO}_2\text{)}_4 = 118.71 + 4(14.01 + 2(16.00)) = 118.71 + 4(14.01 + 32.00) = 118.71 + 184.04 = 302.75 \, \text{g/mol}
] - Molar mass of Pt3N4:
[
\text{Pt} = 195.08 \, \text{g/mol}, \, \text{N} = 14.01 \, \text{g/mol}
]
[
\text{Molar mass of Pt}_3\text{N}_4 = 3(195.08) + 4(14.01) = 585.24 + 56.04 = 641.28 \, \text{g/mol}
]
Step 3: Convert grams to moles
Now, convert the given masses of the reactants to moles:
- Moles of Sn(NO2)4:
[
\text{moles of Sn(NO}_2\text{)}_4 = \frac{102 \, \text{g}}{302.75 \, \text{g/mol}} = 0.337 \, \text{mol}
] - Moles of Pt3N4:
[
\text{moles of Pt}_3\text{N}_4 = \frac{80.2 \, \text{g}}{641.28 \, \text{g/mol}} = 0.125 \, \text{mol}
]
Step 4: Determine the limiting reagent
From the balanced equation, we see that 3 moles of Sn(NO2)4 react with 4 moles of Pt3N4. To determine the limiting reagent, we need to compare the ratio of moles of each reactant to the coefficients in the balanced equation.
- The ratio of Sn(NO2)4 to Pt3N4 should be (3:4). Therefore, for 0.337 mol of Sn(NO2)4, the required amount of Pt3N4 is:
[
\frac{4}{3} \times 0.337 \, \text{mol} = 0.449 \, \text{mol of Pt}_3\text{N}_4
]
But we only have 0.125 mol of Pt3N4, which is less than the required amount. Therefore, Pt3N4 is the limiting reagent.
Step 5: Calculate the theoretical yield
Now, use the amount of the limiting reagent (Pt3N4) to calculate the theoretical yield of Sn3N4. From the balanced equation, 4 moles of Pt3N4 produce 1 mole of Sn3N4. Therefore, 0.125 mol of Pt3N4 will produce:
[
\frac{1}{4} \times 0.125 \, \text{mol} = 0.03125 \, \text{mol of Sn}_3\text{N}_4
]
Now, convert this to grams using the molar mass of Sn3N4:
[
\text{Molar mass of Sn}_3\text{N}_4 = 3(118.71) + 4(14.01) = 356.13 + 56.04 = 412.17 \, \text{g/mol}
]
[
\text{Mass of Sn}_3\text{N}_4 = 0.03125 \, \text{mol} \times 412.17 \, \text{g/mol} = 12.88 \, \text{g}
]
Thus, the theoretical yield of Sn3N4 is 12.88 grams.
Step 6: Calculate the percent yield
Finally, calculate the percent yield using the actual yield (40.2 g) and the theoretical yield (12.88 g):
[
\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{40.2 \, \text{g}}{12.88 \, \text{g}} \times 100 = 312.6\%
]
This indicates that there may be an error in the experimental setup, as a yield above 100% is unusual and suggests impurities or inaccuracies in measurement.
Conclusion:
- Limiting reagent: Pt3N4
- Percent yield: 312.6% (likely indicating experimental error or impurities).