You are given the balanced chemical equation:

1 answer:

5 0

Since there are given data of two reagents, we need to obtain which of reagents are surplus.

0.3618 mol C4H4 - x mol O2
1 mol C4H4 - 5 mol O2

$x = \frac{0.3618 \times 5}{1} = 1.809$
that means that there are too much O2 gas (by 0.009 mol)

so we continue counting using C4H4 data

0.3618 mol C4H4 - x mol H2O
1 mol C4H4 - 2 mol H2O

$x = \frac{0.3618 \times 2}{1} = 0.7236 \: mol$
Mw(H2O)=2×1+16=18 g/mol

n=m/Mw

m=n×Mw

$m = 0.7236 \times 18 = 13.0248 \: g$
answer:13.0248 grams of H2O

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