Multiple Choice
In the reaction 2 C2H6O + 2 K2Cr2O7 + 8 H2SO4 → 2 Cr2(SO4)3 + 2 K2SO4 + 3 CH3COOH + 11 H2O, if the rate of appearance of Cr2(SO4)3 is 1.24 mol/min, what is the rate of disappearance of C2H6O?
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3. Chemical Reactions
Stoichiometry
Multiple Choice
What is the maximum amount of water (in grams) that can be produced when 964.91 g of C3H8 reacts with 511.39 g of oxygen according to the balanced equation: C3H8 + 5 O2 -> 3 CO2 + 4 H2O?
A
327.68 g
B
400.00 g
C
600.00 g
D
500.00 g
Verified step by step guidance1
First, identify the balanced chemical equation for the reaction: \( \text{C}_3\text{H}_8 + 5 \text{O}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} \). This equation shows the stoichiometry of the reactants and products.
Calculate the molar mass of \( \text{C}_3\text{H}_8 \) (propane) and \( \text{O}_2 \) (oxygen). The molar mass of \( \text{C}_3\text{H}_8 \) is \( 3 \times 12.01 + 8 \times 1.01 = 44.11 \text{ g/mol} \). The molar mass of \( \text{O}_2 \) is \( 2 \times 16.00 = 32.00 \text{ g/mol} \).
Convert the mass of \( \text{C}_3\text{H}_8 \) and \( \text{O}_2 \) to moles using their respective molar masses. For \( \text{C}_3\text{H}_8 \), use \( \frac{964.91 \text{ g}}{44.11 \text{ g/mol}} \). For \( \text{O}_2 \), use \( \frac{511.39 \text{ g}}{32.00 \text{ g/mol}} \).
Determine the limiting reactant by comparing the mole ratio from the balanced equation. The balanced equation requires 1 mole of \( \text{C}_3\text{H}_8 \) to react with 5 moles of \( \text{O}_2 \). Calculate the required moles of \( \text{O}_2 \) for the available moles of \( \text{C}_3\text{H}_8 \) and compare it with the available moles of \( \text{O}_2 \).
Using the limiting reactant, calculate the maximum moles of \( \text{H}_2\text{O} \) produced. From the balanced equation, 1 mole of \( \text{C}_3\text{H}_8 \) produces 4 moles of \( \text{H}_2\text{O} \). Convert the moles of \( \text{H}_2\text{O} \) to grams using the molar mass of water (\( 18.02 \text{ g/mol} \)).
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