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Answer :
Final answer:
The yield percent for producing 190 g of N2O4 gas from the reaction of 84 g of Nitrogen gas with 128 g of Oxygen gas can be calculated using the given information and the stoichiometry of the reaction. The yield percent is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
Explanation:
To calculate the yield percent, we first need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's calculate the moles of Nitrogen gas and Oxygen gas:
- Moles of Nitrogen gas = given mass of Nitrogen gas / molar mass of Nitrogen gas
- Moles of Oxygen gas = given mass of Oxygen gas / molar mass of Oxygen gas
Next, we need to determine the stoichiometry of the reaction. The balanced chemical equation for the reaction is:
2 N2 + O2 → 2 N2O4
From the balanced equation, we can see that 2 moles of Nitrogen gas react with 1 mole of Oxygen gas to produce 2 moles of N2O4 gas.
Now, we can calculate the theoretical yield of N2O4 gas:
- Moles of N2O4 gas = (moles of Nitrogen gas / 2) * (molar mass of N2O4 gas)
Finally, we can calculate the yield percent:
- Yield percent = (actual yield / theoretical yield) * 100
Learn more about calculating yield percent in a chemical reaction here:
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