To determine how many grams of oxygen gas (O₂) are required to fully combust 7.33 moles of acetylene (C₂H₂), we need to start with the balanced chemical equation for the combustion of acetylene:
\[ 2 \text{C}_2\text{H}_2 + 5 \text{O}_2 \rightarrow 4 \text{CO}_2 + 2 \text{H}_2\text{O} \]
From the balanced equation, we can see that 2 moles of acetylene (C₂H₂) require 5 moles of oxygen (O₂) for complete combustion.
First, we can find out how many moles of oxygen are needed for 7.33 moles of acetylene:
Using the stoichiometric ratio from the equation: \[ \text{Moles of O₂ required} = (7.33 \text{ moles C₂H₂}) \times \left(\frac{5 \text{ moles O₂}}{2 \text{ moles C₂H₂}}\right) = 7.33 \times 2.5 = 18.325 \text{ moles O₂} \]
Next, we need to convert moles of oxygen to grams. To do this, we use the molar mass of oxygen gas (O₂). The molar mass of O₂ is approximately 32.00 g/mol.
Now, we can calculate the mass of oxygen gas required:
\[ \text{Mass of O₂} = \text{moles of O₂} \times \text{molar mass of O₂} \] \[ \text{Mass of O₂} = 18.325 \text{ moles O₂} \times 32.00 \text{ g/mol} = 585.04 \text{ grams O₂} \]
Thus, the amount of oxygen gas required to fully combust 7.33 moles of acetylene is approximately 585.04 grams.
1 answer