How man grams of nitrogen monoxide are needed to react with adequate oxygen to produce 89.6 L of nitrogen dioxide
120.g NO
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To determine the grams of nitrogen monoxide (NO) needed to react with oxygen (O₂) to produce a given volume of nitrogen dioxide (NO₂), we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction.
2NO + O₂ → 2NO₂
Step 2: Determine the molar ratio between NO and NO₂.
From the balanced chemical equation, we can see that 2 moles of NO react to form 2 moles of NO₂.
Step 3: Convert the volume of NO₂ to moles.
We need to convert the given volume of NO₂ (89.6 L) into moles using the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the moles, R is the gas constant, and T is the temperature. Assuming standard conditions (1 atm and 273 K), the volume can be directly converted to moles:
n(NO₂) = V(NO₂) / Vm
where Vm is the molar volume of a gas at STP.
Step 4: Apply the molar ratio to determine the moles of NO.
Since the molar ratio is 2:2, the moles of NO are the same as the moles of NO₂, so:
n(NO) = n(NO₂)
Step 5: Convert moles of NO to grams.
Use the molar mass of nitrogen monoxide (NO) to convert moles to grams:
mass(NO) = n(NO) × molar mass(NO)
By following these steps and performing the calculations, you can find the amount of nitrogen monoxide needed in grams to react with adequate oxygen to produce 89.6 L of nitrogen dioxide.