Answer
27.5 g of $CO_2$ are produced from 32.0 g of $O_2$.
Work Step by Step
1. Calculate the molar mass of $O_2$:
$O: 16.00g * 2= 32.00g $
$ \frac{1 mole (O_2)}{ 32.00g (O_2)}$ and $ \frac{ 32.00g (O_2)}{1 mole (O_2)}$
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 $O_2$ to $CO_2$ is 8 to 5:
$ \frac{ 5 moles(CO_2)}{ 8 moles (O_2)}$ and $ \frac{ 8 moles (O_2)}{ 5 moles(CO_2)}$
3. Calculate the molar mass for $CO_2$:
$C: 12.01g $
$O: 16.00g * 2= 32.00g $
12.01g + 32.00g = 44.01g
$ \frac{1 mole (CO_2)}{ 44.01g (CO_2)}$ and $ \frac{ 44.01g (CO_2)}{1 mole (CO_2)}$
4. Use the conversion factors to find the mass of $CO_2$
$32.0g(O_2) \times \frac{1 mole(O_2)}{ 32.00g( O_2)} \times \frac{ 5 moles(CO_2)}{ 8 moles (O_2)} \times \frac{ 44.01 g (CO_2)}{ 1 mole (CO_2)} = 27.5g (CO_2)$
