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Answer :
Final answer:
The limiting reagent in the reaction between 25.0 g of CO and 6.00 g of H2 to form methanol (CH3OH) is CO because there are fewer moles of CO than required for the stoichiometry of the reaction.
Explanation:
To find the limiting reagent, we first need to express the given masses of CO and H2 in terms of moles for the reaction CO + 2H2 -> CH3OH. For CO, we have 25.0 g / 28.01 g/mol = 0.892 mol CO. For H2, we have 6.00 g / 2.02 g/mol = 2.97 mol H2. The stoichiometry of the reaction requires 2 moles of H2 for each mole of CO. But we have more than 2 moles of H2 for each mole of CO (2.97/0.892 = 3.33). Therefore, CO is the limiting reagent and H2 is in excess.
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