Respuesta :
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
[tex]x = \frac{0.3618 \times 5}{1} = 1.809[/tex]
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
[tex]x = \frac{0.3618 \times 2}{1} = 0.7236 \: mol[/tex]
Mw(H2O)=2×1+16=18 g/mol
n=m/Mw
m=n×Mw
[tex]m = 0.7236 \times 18 = 13.0248 \: g[/tex]
answer:13.0248 grams of H2O
0.3618 mol C4H4 - x mol O2
1 mol C4H4 - 5 mol O2
[tex]x = \frac{0.3618 \times 5}{1} = 1.809[/tex]
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
[tex]x = \frac{0.3618 \times 2}{1} = 0.7236 \: mol[/tex]
Mw(H2O)=2×1+16=18 g/mol
n=m/Mw
m=n×Mw
[tex]m = 0.7236 \times 18 = 13.0248 \: g[/tex]
answer:13.0248 grams of H2O
The maximum mass of water that could be produced is 13.0 g
Stoichiometry
From the question, we are to determine the maximum mass of water that could be produced.
From the given balanced chemical equation,
C₄H₄ + 5O₂ → 4CO₂ + 2H₂O
This means, 1 mole of Câ‚„Hâ‚„ reacts with 5 moles of Oâ‚‚ to produce 4 moles of COâ‚‚ and 2 moles of Hâ‚‚O
If 1 mole of Câ‚„Hâ‚„ reacts with 5 moles of Oâ‚‚ to produce 2 moles of Hâ‚‚O
Then,
0.3618 mole of Câ‚„Hâ‚„ will react with 1.809 moles of Oâ‚‚ to produce 0.7236 mole of Hâ‚‚O
Therefore, the number of moles of water that could be produced is 0.7236 mole
Now, for the mass of water that could be produced
Using the formula,
Mass = Number of moles × Molar mass
Molar mass of Hâ‚‚O = 18.015 g/mol
Therefore,
Mass of water produced = 0.7236 × 18.015
Mass of water produced = 13.035654 g
Mass of water produced ≅ 13.0 g
Hence, the maximum mass of water that could be produced is 13.0 g
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