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According to the following reaction, what mass of PbCl2 can form from 235 mL of 0.110 M KCl solution? Assume that there is excess Pb(NO3)2.
2 KCl(aq) + Pb(NO3)2(aq) ⟶ PbCl2(s) + 2 KNO3(aq)
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3. Chemical Reactions
Stoichiometry
Scelta multipla
In the reaction 2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g), if 0.300 mol of octane (C8H18) is allowed to react with 0.870 mol of oxygen (O2) and oxygen is the limiting reactant, resulting in the production of 0.626 mol of water (H2O), how much octane is left unreacted?
A
0.174 mol
B
0.100 mol
C
0.200 mol
D
0.126 mol
Guida verificata passo dopo passo1
First, identify the balanced chemical equation for the reaction: 2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g). This equation tells us the stoichiometric ratios between the reactants and products.
Determine the stoichiometric ratio between octane (C8H18) and oxygen (O2) from the balanced equation. For every 2 moles of C8H18, 25 moles of O2 are required. Therefore, the ratio is 2:25.
Calculate the amount of octane that reacts with the given amount of oxygen. Since oxygen is the limiting reactant, use the stoichiometric ratio to find out how much octane reacts with 0.870 mol of O2. Use the formula: \( \text{moles of C8H18 reacted} = \frac{2}{25} \times \text{moles of O2} \).
Subtract the moles of octane that reacted from the initial moles of octane to find the moles of octane left unreacted. Use the formula: \( \text{moles of C8H18 unreacted} = \text{initial moles of C8H18} - \text{moles of C8H18 reacted} \).
Verify the calculation by checking the stoichiometry of the reaction and ensuring that the amount of water produced (0.626 mol) aligns with the stoichiometric predictions based on the limiting reactant.
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