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To solve the question of how many moles of [tex]\( KMnO_4 \)[/tex] are consumed to form 0.886 moles of carbon dioxide, let's go through the problem step-by-step using stoichiometry based on the chemical equation provided:
[tex]\[ 14 \, KMnO_4 + 4 \, C_3H_5(OH)_3 \rightarrow 7 \, K_2CO_3 + 7 \, Mn_2O_3 + 5 \, CO_2 + 16 \, H_2O \][/tex]
Step-by-step solution:
1. Identify the relationship between [tex]\( KMnO_4 \)[/tex] and [tex]\( CO_2 \)[/tex] in the balanced equation:
From the balanced chemical equation, we know that 14 moles of [tex]\( KMnO_4 \)[/tex] produce 5 moles of [tex]\( CO_2 \)[/tex].
2. Set up a proportional relationship:
To determine how many moles of [tex]\( KMnO_4 \)[/tex] are needed to produce 0.886 moles of [tex]\( CO_2 \)[/tex], we can set up the following proportion:
[tex]\[ \frac{\text{Moles of } KMnO_4}{14} = \frac{0.886 \text{ moles of } CO_2}{5} \][/tex]
3. Solve for the moles of [tex]\( KMnO_4 \)[/tex]:
Now, cross-multiply and solve for the moles of [tex]\( KMnO_4 \)[/tex]:
[tex]\[ \text{Moles of } KMnO_4 = \frac{0.886 \times 14}{5} \][/tex]
4. Calculate the result:
[tex]\[ \text{Moles of } KMnO_4 = \frac{12.404}{5} = 2.4808 \][/tex]
Therefore, 2.4808 moles of [tex]\( KMnO_4 \)[/tex] are consumed to form 0.886 moles of carbon dioxide.
[tex]\[ 14 \, KMnO_4 + 4 \, C_3H_5(OH)_3 \rightarrow 7 \, K_2CO_3 + 7 \, Mn_2O_3 + 5 \, CO_2 + 16 \, H_2O \][/tex]
Step-by-step solution:
1. Identify the relationship between [tex]\( KMnO_4 \)[/tex] and [tex]\( CO_2 \)[/tex] in the balanced equation:
From the balanced chemical equation, we know that 14 moles of [tex]\( KMnO_4 \)[/tex] produce 5 moles of [tex]\( CO_2 \)[/tex].
2. Set up a proportional relationship:
To determine how many moles of [tex]\( KMnO_4 \)[/tex] are needed to produce 0.886 moles of [tex]\( CO_2 \)[/tex], we can set up the following proportion:
[tex]\[ \frac{\text{Moles of } KMnO_4}{14} = \frac{0.886 \text{ moles of } CO_2}{5} \][/tex]
3. Solve for the moles of [tex]\( KMnO_4 \)[/tex]:
Now, cross-multiply and solve for the moles of [tex]\( KMnO_4 \)[/tex]:
[tex]\[ \text{Moles of } KMnO_4 = \frac{0.886 \times 14}{5} \][/tex]
4. Calculate the result:
[tex]\[ \text{Moles of } KMnO_4 = \frac{12.404}{5} = 2.4808 \][/tex]
Therefore, 2.4808 moles of [tex]\( KMnO_4 \)[/tex] are consumed to form 0.886 moles of carbon dioxide.
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