Given the following equation: 4Na(s) + O2(g ) → 2Na2O(s) How many L of O2 gas at STP, are needed to react with 55.0 g of Na?
The correct answer and explanation is:
To solve this, we need to follow the steps of stoichiometry. First, we need to determine the moles of sodium (Na) involved in the reaction, then use the stoichiometric relationship between sodium and oxygen to find the volume of O2 needed at STP.
Step 1: Convert grams of Na to moles
The molar mass of sodium (Na) is approximately 22.99 g/mol. So, the number of moles of Na in 55.0 g is: moles of Na=55.0 g22.99 g/mol=2.39 mol\text{moles of Na} = \frac{55.0 \, \text{g}}{22.99 \, \text{g/mol}} = 2.39 \, \text{mol}
Step 2: Use the balanced equation to find the ratio of Na to O2
From the balanced chemical equation: 4Na(s)+O2(g)→2Na2O(s)4Na(s) + O2(g) \rightarrow 2Na2O(s)
The ratio of Na to O2 is 4:1. This means for every 4 moles of Na, 1 mole of O2 is required.
Now, we calculate the moles of O2 needed to react with 2.39 mol of Na: moles of O2=2.39 mol Na4=0.5975 mol O2\text{moles of O2} = \frac{2.39 \, \text{mol Na}}{4} = 0.5975 \, \text{mol O2}
Step 3: Convert moles of O2 to volume at STP
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 L. Therefore, the volume of O2 needed is: volume of O2=0.5975 mol×22.4 L/mol=13.4 L\text{volume of O2} = 0.5975 \, \text{mol} \times 22.4 \, \text{L/mol} = 13.4 \, \text{L}
Final Answer:
The volume of O2 gas required to react with 55.0 g of Na is 13.4 L.
This calculation shows how to use stoichiometry to relate the mass of one reactant (Na) to the volume of another reactant (O2) at STP, ensuring that all units cancel appropriately and the final result is in the correct form.