Answer:
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<em>a) Balanced chemical equation:</em>
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<em>b) Theoretical yield:</em>
c) % yield:
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>
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<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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The percent composition of hydrogen is 18.3%.
What is percent composition?
The term percent composition refers to the percentage of a particular component in a compound. It is contained as the ratio of the mass of that component to the total mass multiplied by 100.
From the law of conservation of mass, total mass of beryllium hydride(BeH2) = 85.2 g
Mass of hydrogen = 15.6 g
Percent composition of hydrogen = 15.6 g/ 85.2 g × 100/1 = 18.3%
The chemical reaction in which atoms are changed
Answer:
the concentration of hydrogen iodide will be higher than it was in the original equilibrium conditions.
Answer:
704.6 g CO2
Explanation:
MM sucrose = 342.3 g/mol
MM CO2 = 44.01 g/mol
g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2