Question

The combustion of propane may be described by the chemical equation,
C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(g)
How many grams of O2(g) are needed to completely burn 97.3 g of C3H8(g)?

Answer 1

Answer on Question #40464, Chemistry, Other

Question

The combustion of propane may be described by the chemical equation,

$$\mathrm{C_3H_8(g)} + 5\mathrm{O_2(g)} \rightarrow 3\mathrm{CO_2(g)} + 4\mathrm{H_2O(g)}$$

How many grams of $\mathrm{O_2(g)}$ are needed to completely burn $97.3\,\mathrm{g}$ of $\mathrm{C_3H_8(g)}$?

Answer

$$\begin{array}{l} \mathrm{M}(\mathrm{C_3H_8}) = 44\,\mathrm{g/mol} \\ \mathrm{n}(\mathrm{C_3H_8}) = \mathrm{m}(\mathrm{C_3H_8})/\mathrm{M}(\mathrm{C_3H_8}) = 97.3\,\mathrm{g} / 44\,\mathrm{g/mol} = 2.21\,\mathrm{mol} \\ \mathrm{n}(\mathrm{O_2}) = 5 \cdot \mathrm{n}(\mathrm{C_3H_8}) = 5 \cdot 2.21 = 11.05\,\mathrm{mol} \\ \mathrm{M}(\mathrm{O_2}) = 32\,\mathrm{g/mol} \\ \mathrm{m}(\mathrm{O_2}) = \mathrm{n}(\mathrm{O_2}) \cdot \mathrm{M}(\mathrm{O_2}) = 11.05\,\mathrm{mol} \cdot 32\,\mathrm{g/mol} = 353.6\,\mathrm{g} \\ \end{array}$$

Answer: 353.6 g