Answer
a. $73.9g (C_4H_6O_3)$
b. $ 130g (C_9H_8O_4)$
Work Step by Step
a.
1. Calculate the number of moles of $C_7H_6O_3$:
12.01* 7 + 1.008* 6 + 16.00* 3 = 138.12g/mol
$1.00 \times 10^2 g \times \frac{1 mol}{ 138.12g} = 0.724mol (C_7H_6O_3)$
According to the balanced reaction:
The ratio of $C_7H_6O_3$ to $C_4H_6O_3$ is 1 to 1:
$0.724 mol (C_7H_6O_3) \times \frac{ 1 mol(C_4H_6O_3)}{ 1 mol (C_7H_6O_3)} = 0.724mol (C_4H_6O_3)$
2. Calculate the mass of $C_4H_6O_3$:
12.01* 4 + 1.008* 6 + 16.00* 3 = 102.09g/mol
$0.724 mol \times \frac{ 102.09 g}{ 1 mol} = 73.9g (C_4H_6O_3)$
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b.
The ratio of $C_7H_6O_3$ to $C_9H_8O_4$ is 1 to 1:
$0.724 mol (C_7H_6O_3) \times \frac{ 1 mol(C_9H_8O_4)}{ 1 mol (C_7H_6O_3)} = 0.724mol (C_9H_8O_4)$
2. Calculate the mass of $C_9H_8O_4$:
12.01* 9 + 1.008* 8 + 16.00* 4 = 180.15g/mol
$0.724 mol \times \frac{ 180.15 g}{ 1 mol} = 130g (C_9H_8O_4)$
