Answer
$242$ g of glucose $(C_6H_{12}O_6)$ are required to form 124 g of ethanol.
Work Step by Step
1. Calculate the molar mass of $C_2H_6O$:
$C: 12.01g * 2= 24.02g $
$H: 1.008g * 6= 6.048g $
$O: 16.00g$
24.02g + 6.048g + 16.00g = 46.07g
$ \frac{1 mole (C_2H_6O)}{ 46.07g (C_2H_6O)}$ and $ \frac{ 46.07g (C_2H_6O)}{1 mole (C_2H_6O)}$
2. The balanced reaction is:
$C_6H_{12}O_6 --\gt 2C_2H_6O + 2CO_2$
According to the coefficients, the ratio of $C_2H_6O$ to $C_6H_{12}O_6$ is 2 to 1:
$ \frac{ 1 mole(C_6H_{12}O_6)}{ 2 moles (C_2H_6O)}$ and $ \frac{ 2 moles (C_2H_6O)}{ 1 mole(C_6H_{12}O_6)}$
3. Calculate the molar mass for $C_6H_{12}O_6$:
$C: 12.01g * 6= 72.06g $
$H: 1.008g * 12= 12.10g $
$O: 16.00g * 6= 96.00g $
72.06g + 12.10g + 96.00g = 180.16g
$ \frac{1 mole (C_6H_{12}O_6)}{ 180.156g (C_6H_{12}O_6)}$ and $ \frac{ 180.156g (C_6H_{12}O_6)}{1 mole (C_6H_{12}O_6)}$
4. Use the conversion factors to find the mass of $C_6H_{12}O_6$
$124g(C_2H_6O) \times \frac{1 mole(C_2H_6O)}{ 46.07g( C_2H_6O)} \times \frac{ 1 mole(C_6H_{12}O_6)}{ 2 moles (C_2H_6O)} \times \frac{ 180.16 g (C_6H_{12}O_6)}{ 1 mole (C_6H_{12}O_6)} = 242g (C_6H_{12}O_6)$