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andrew11 [14]
3 years ago
15

Given the following reaction:

Chemistry
1 answer:
mojhsa [17]3 years ago
4 0

Answer:

Option D. 17.5

Explanation:

Equiibrium is: CO + 2H₂  ⇄  CH₃OH

1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.

Initially we have 0.42 moles of CO and 0.42 moles of H₂

If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.

So in the equilibrium we may have:

0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂

Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen

Finally 0.13 moles of methanol, are found after the equilibrium reach the end.

Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²

0.13 / (0.29 . 0.16²)

Kc = 17.5

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Consider the balanced equation for the following reaction:
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<u>Answer:</u> The amount of carbon dioxide formed in the reaction is 5.663 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of oxygen gas = 8 g

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Molar mass of carbon dioxide = 44 g/mol

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Putting values in equation 1, we get:

0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g

To calculate the experimental yield of carbon dioxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

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Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

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3 years ago
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