Answer to Question #3142 in Organic Chemistry for mellisa
?CH3OH + ?O2 => ?HCO2H + ?H2O,& & what is the maximum amount of HCO2H(46.0254 g/mol) which could be formed from
10.61 mol of CH3OH (32.0419 g/mol) and13.09 mol of O2 (31.9988 g/mol)?Answer in units of g.
CH3OH + O2 --> HCO2H + H2O
so, only 10.61 mol of CH3OH could react with 10.61 mol of O2.
ν(HCO2H) = ν(O2) = 10.61 mol.
m(HCO2H) = 46.0254 [g/mol] * 10.61 mol = 488.3250 g.
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