What volume of carbon monoxide gas (at STP) is needed when 1.31 moles of oxygen gas react completely in the following equation?
2 CO + O2 --> 2 CO2
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To find the volume of carbon monoxide gas needed, we can use the balanced equation and the ideal gas law. Here are the steps you can follow to solve the problem:
Step 1: Write the balanced equation:
2 CO + O2 → 2 CO2
Step 2: Determine the stoichiometry of the reaction.
From the balanced equation, we can see that for every 2 moles of carbon monoxide (CO) that react, 1 mole of oxygen gas (O2) is required to produce 2 moles of carbon dioxide (CO2).
Step 3: Calculate the moles of carbon monoxide needed.
Since we have 1.31 moles of oxygen gas, we can use the stoichiometry from Step 2 to determine the moles of carbon monoxide needed. From the stoichiometry, we know that 1 mole of oxygen gas is required for every 2 moles of carbon monoxide. Therefore, to find the moles of carbon monoxide needed, we can use the following ratio:
1.31 moles O2 * (2 moles CO / 1 mole O2) = 2.62 moles CO
Step 4: Use the ideal gas law to find the volume of carbon monoxide.
The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin. At STP (standard temperature and pressure), the temperature is 273.15 K, and the pressure is 1 atm.
Plugging in the values into the ideal gas law equation:
V = (n * R * T) / P
V = (2.62 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
V ≈ 58.76 L
Therefore, approximately 58.76 liters of carbon monoxide gas is needed when 1.31 moles of oxygen gas react completely in the given reaction at STP.