contestada

You are given the balanced chemical equation:
C4H4 + 5O2 4CO2 + 2H2O.
If 0.3618 moles of C4H4 are allowed to react with 1.818 moles of O2, and this is the only reaction which occurs, what is the maximum mass of water that could be produced?

Respuesta :

Awste
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

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 C₄H₄ reacts with 5 moles of O₂ to produce 4 moles of CO₂ and 2 moles of H₂O

If 1 mole of C₄H₄ reacts with 5 moles of O₂ to produce 2 moles of H₂O

Then,

0.3618 mole of C₄H₄ will react with 1.809 moles of O₂ to produce 0.7236 mole of H₂O

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 H₂O = 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

Learn more on Stoichiometry here: https://brainly.com/question/26545586

Otras preguntas