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How many grams of hydrogen are needed to produce 68.00 moles of methanol?
- Roger the MoleLv 72 months ago
Supposing all the hydrogen is incorporated into the methanol, and that there is no hydrogen in the carbon-containing reactant(s):
(68.00 mol CH3OH) x (4 mol H / 1 mol CH3OH) x (1.007947 g H/mol) = 274.2 g H