Question

excuse me!! mr. saquib appko jo bolna hai bol do kyuki main appko tab like Karti thi aab nhi ok and I hate u.​

Answer 1

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Answer 2

Answer:

Well, we need a stoichiometrically balanced equation to represent the combustion … and as always, we (i) balance the carbons as carbon dioxide; (ii) balance the hydrogens as water; and (iii) balance the oxygens as dioxygen gas…

H3COH(l)+32O2(g)⟶CO2(g)+2H2O(l)

And thus we get 2 equiv water per equiv methanol combusted…

nMeOH=209∙g32.04∙g∙mol−1=6.52∙mol

And so we need 32 equiv of dioxygen gas, i.e. 9.78∙mol , a mass of approx 313∙g . And we get approx 13∙mol of water … i.e. a mass of 235∙g

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