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Given:
[tex]CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O[/tex]

In the chemical reaction, if 10 moles of [tex]H_2O[/tex] are produced, [tex]\square[/tex] moles of [tex]CO_2[/tex] are also produced.

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To determine how many moles of CO[tex]\(_2\)[/tex] are produced when 10 moles of H[tex]\(_2\)[/tex]O are produced, we can use the stoichiometric coefficients from the balanced chemical equation:

[tex]\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \][/tex]

From the balanced equation, we see that:
- 1 mole of CH[tex]\(_4\)[/tex] produces 1 mole of CO[tex]\(_2\)[/tex]
- 2 moles of O[tex]\(_2\)[/tex] are used to produce 1 mole of CO[tex]\(_2\)[/tex]
- To each mole of CO[tex]\(_2\)[/tex], 2 moles of H[tex]\(_2\)[/tex]O are produced

Given that there are 10 moles of H[tex]\(_2\)[/tex]O produced, we use the ratio of H[tex]\(_2\)[/tex]O to CO[tex]\(_2\)[/tex]:

[tex]\[ \text{2 moles of H}_2\text{O} \longrightarrow \text{1 mole of CO}_2 \][/tex]

Since we have 10 moles of H[tex]\(_2\)[/tex]O, we divide by 2 to find the corresponding moles of CO[tex]\(_2\)[/tex]:

[tex]\[ \frac{10 \text{ moles of H}_2\text{O}}{2} = 5 \text{ moles of CO}_2 \][/tex]

Therefore, [tex]\(\boxed{5}\)[/tex] moles of CO[tex]\(_2\)[/tex] are also produced.
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