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Answer :
Sure! Let's go through the steps to determine how many grams of water are produced from 6.00 moles of hydrogen in the reaction [tex]\(2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2 \text{O}\)[/tex].
1. Identify the Mole Ratio: In the balanced chemical equation, [tex]\(2 \text{H}_2\)[/tex] produces [tex]\(2 \text{H}_2 \text{O}\)[/tex]. This shows that 2 moles of hydrogen gas produce 2 moles of water, or a 1:1 ratio between moles of [tex]\(\text{H}_2\)[/tex] and [tex]\(\text{H}_2\text{O}\)[/tex].
2. Calculate Moles of Water Produced: Since the ratio is 1:1, 6.00 moles of [tex]\(\text{H}_2\)[/tex] will produce 6.00 moles of [tex]\(\text{H}_2\text{O}\)[/tex].
3. Find the Molar Mass of Water: The molar mass of water ([tex]\(\text{H}_2\text{O}\)[/tex]) is approximately 18.01528 grams per mole.
4. Calculate Grams of Water Produced: Multiply the moles of water by the molar mass of water:
[tex]\[
\text{Grams of } \text{H}_2\text{O} = 6.00 \text{ mol} \times 18.01528 \text{ g/mol} = 108.09168 \text{ g}
\][/tex]
Rounding to appropriate significant figures (based on the 6.00 moles given in the problem), the answer is approximately [tex]\(108 \text{ g}\)[/tex].
So, the correct answer is d) 108 g.
1. Identify the Mole Ratio: In the balanced chemical equation, [tex]\(2 \text{H}_2\)[/tex] produces [tex]\(2 \text{H}_2 \text{O}\)[/tex]. This shows that 2 moles of hydrogen gas produce 2 moles of water, or a 1:1 ratio between moles of [tex]\(\text{H}_2\)[/tex] and [tex]\(\text{H}_2\text{O}\)[/tex].
2. Calculate Moles of Water Produced: Since the ratio is 1:1, 6.00 moles of [tex]\(\text{H}_2\)[/tex] will produce 6.00 moles of [tex]\(\text{H}_2\text{O}\)[/tex].
3. Find the Molar Mass of Water: The molar mass of water ([tex]\(\text{H}_2\text{O}\)[/tex]) is approximately 18.01528 grams per mole.
4. Calculate Grams of Water Produced: Multiply the moles of water by the molar mass of water:
[tex]\[
\text{Grams of } \text{H}_2\text{O} = 6.00 \text{ mol} \times 18.01528 \text{ g/mol} = 108.09168 \text{ g}
\][/tex]
Rounding to appropriate significant figures (based on the 6.00 moles given in the problem), the answer is approximately [tex]\(108 \text{ g}\)[/tex].
So, the correct answer is d) 108 g.
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