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Which conversion would show how many moles of water could be formed from 3.50 moles of oxygen in the following equation?

[tex]
C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O
[/tex]

A. [tex] 8 \text{ moles } O_2 \times \frac{6 \text{ moles } H_2O}{3.5 \text{ moles } O_2} [/tex]

B. [tex] 3.50 \text{ moles } O_2 \times \frac{6 \text{ moles } H_2O}{8 \text{ moles } O_2} [/tex]

C. [tex] 8 \text{ moles } H_2O \times \frac{6 \text{ moles } H_2O}{8 \text{ moles } O_2} [/tex]

D. [tex] 6 \text{ moles } O_2 \times \frac{3.5 \text{ moles } H_2O}{8 \text{ moles } O_2} [/tex]


Sagot :

To determine how many moles of water could be formed from 3.50 moles of oxygen using the given chemical equation:

[tex]\[ \mathrm{C_5H_{12} + 8 \, O_2 \rightarrow 5 \, CO_2 + 6 \, H_2O} \][/tex]

we need to use the stoichiometric relationships from the balanced equation. The balanced equation states that 8 moles of [tex]\( \mathrm{O_2} \)[/tex] produce 6 moles of [tex]\( \mathrm{H_2O} \)[/tex].

Let's convert the given amount of [tex]\( \mathrm{O_2} \)[/tex] to moles of [tex]\( \mathrm{H_2O} \)[/tex]:

Given moles of [tex]\( \mathrm{O_2} \)[/tex] = 3.50 moles

Using the ratio from the balanced equation:
[tex]\[ \text{Moles of } \mathrm{H_2O} = \text{Moles of } \mathrm{O_2} \times \frac{6 \text{ moles } \mathrm{H_2O}}{8 \text{ moles } \mathrm{O_2}} \][/tex]

So,
[tex]\[ 3.50 \text{ moles } \mathrm{O_2} \times \frac{6 \text{ moles } \mathrm{H_2O}}{8 \text{ moles } \mathrm{O_2}} \][/tex]

Upon calculating this, we find:
[tex]\[ 3.50 \text{ moles } \mathrm{O_2} \times 0.75 = 2.625 \text{ moles } \mathrm{H_2O} \][/tex]

Thus, 3.50 moles of [tex]\( \mathrm{O_2} \)[/tex] would produce 2.625 moles of [tex]\( \mathrm{H_2O} \)[/tex].

Based on this calculation, the correct conversion from the given options is:

[tex]\[ 3.50 \text{ moles } O_2 \times \frac{6 \text{ moles } H_2O }{8 \text{ moles } O_2} \][/tex]

This step-by-step solution shows that the second option matches the proper conversion to determine how many moles of water would be formed.
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