Answer
48 g of $C_5H_{12}$ are needed to produce 72 g of water.
Work Step by Step
1. Calculate the molar mass of $H_2O$:
$H: 1.008g * 2= 2.016g $
$O: 16.00g$
2.016g + 16.00g = 18.02g
$ \frac{1 mole (H_2O)}{ 18.02g (H_2O)}$ and $ \frac{ 18.02g (H_2O)}{1 mole (H_2O)}$
2. The balanced reaction is:
$C_5H_{12}(g) + 8O_2(g) --\gt 5CO_2(g) + 6H_2O(g)$
According to the coefficients, the ratio of $H_2O$ to $C_5H_{12}$ is 6 to 1:
$ \frac{ 1 mole(C_5H_{12})}{ 6 moles (H_2O)}$ and $ \frac{ 6 moles (H_2O)}{ 1 mole(C_5H_{12})}$
3. Calculate the molar mass for $C_5H_{12}$:
Molar mass :
$C: 12.01g * 5= 60.05g $
$H: 1.008g * 12= 12.10g $
60.05g + 12.10g = 72.15g
$ \frac{1 mole (C_5H_{12})}{ 72.15g (C_5H_{12})}$ and $ \frac{ 72.15g (C_5H_{12})}{1 mole (C_5H_{12})}$
4. Use the conversion factors to find the mass of $C_5H_{12}$
$72g(H_2O) \times \frac{1 mole(H_2O)}{ 18.02g( H_2O)} \times \frac{ 1 mole(C_5H_{12})}{ 6 moles (H_2O)} \times \frac{ 72.15 g (C_5H_{12})}{ 1 mole (C_5H_{12})} = 48g (C_5H_{12})$
