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The combustion of 9.8 moles of methanol will produce 352.8 grams of water.
First, let us look at the balanced equation of combustion of methanol:
[tex]2CH_3OH(l)+3O_2(g)-- > 2CO_2(g)+4H_2O(l)[/tex]
The ratio of number of moles of methanol burned and that of water produced is 1:2.
Thus, 9.8 moles of methanol will produce 19.6 moles of water.
Mass of 19.6 moles water = 19.6 x 18 = 352.8 grams
More on stoichiometric calculations can be found here: https://brainly.com/question/27287858
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