Li- outer energy level
1 answer:
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Answer:
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<em>a) Balanced chemical equation:</em>
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<em>b) Theoretical yield:</em>
- 27.4 g of N₂O₃
c) % yield:
- 24.5%
Explanation:
The complete question is:
<em>In a particular reaction 6.80g of dinitrogen trioxide gas (N₂0₃) was actually produced by reacting 8.75g of oxygen gas (O₂) with excess nitrogen gas (N₂)</em>
<em>a) Write a balanced chemical equation for the reaction. Be sure to include physical states in the equation.</em>
<em>b) Calculate the theoretical yield (in grams) of dinitrogen trioxide: Use dimensional analysis</em>
<em>c) Calculate the % yield of the product</em>
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<h2>Solution</h2><em />
<em>a) Write a balanced chemical equation for the reaction. Be sure to include physical states in the equation.</em>
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Check the balance:
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Atom Left-handside Right-hand side
N 2×2=4 2×2=4
O 3×2=6 2×3=6
- Mole ratio: it is the ratio of the coefficients of the balanced equation
<em>b) Calculate the theoretical yield (in grams) of dinitrogen trioxide: Use dimensional analysis</em>
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<u>1. Convert 8.75 g of O₂(g) to number of moles</u>
- number of moles = mass in grams / molar mass
- molar mass of O₂ = 15.999g/mol
- number of moles = 8.75g / 15.999 g/mol = 0.5469 mol O₂
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<u>2. Use dimensional analysis to calculate the maximum number of moles of N₂O₃(g) that can be produced</u>
<u>3. Convert to mass in grams</u>
- mass = number of moles × molar mass
- molar mass of N₂O3 = 76.01g/mol
- mass = 0.3646mol × 76.01g/mol = 27.7g N₂O3
<em>c) Calculate the % yield of the product</em>
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Formula:
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- %yield = (actual yield/theoretical yield)×100
Substitute and compute:
- % yield = (6.80g/27.7g)×100 = 24.5%
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