Question

1. A chemist begins with 10.0 g of Al(NO3)3 and 140 ml of H2SO4.

2. Calculate the theoretical yield of Al2(SO4)3 produced by this reaction, using the equation: Al(NO3)3 + H2SO4 --> Al2(SO4)3 + HNO3.
3. Perform the reaction, collecting 10.25 g of Al2(SO4)3.
4. Calculate the percent yield using the following equation: Percent Yield = (Actual yield/Theoretical yield)*100.

Answer 1

1. To calculate the theoretical yield of Al2(SO4)3, we need to first calculate the moles of Al(NO3)3, which can be done by dividing 10.0 g by the molar mass of Al(NO3)3 which is 133.99 g/mol. This equals 0.0744 mol of Al(NO3)3.

Next, we can calculate the moles of H2SO4 by dividing 140 ml by the density of sulfuric acid (1.84g/ml), and then dividing this answer by 98 g/mol to get 0.7145 mol H2SO4.

Since the equation has a 1:1 ratio of Al(NO3)3 to H2SO4, the moles of Al2(SO4)3 produced will also be 0.7145 mol. Finally, we can multiply this number by the molar mass of Al2(SO4)3 (342.15 g/mol) to get the theoretical yield of Al2(SO4)3 produced, which is 242.5 g.

2. The actual yield of Al2(SO4)3 is 10.25 g.

3. To calculate the percent yield, we use the following equation: Percent Yield = (Actual yield/Theoretical yield)*100. Plugging in our numbers, we get: Percent Yield = (10.25/242.5)*100 = 4.23%.