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erma4kov [3.2K]
3 years ago
10

What poisonous gas is found in the exhaust fumes of car engines?

Chemistry
2 answers:
Xelga [282]3 years ago
6 0
The poisonous gases found in the exhaust fumes of car engines are the products of combustion which are carbon dioxide and carbon monoxide. They are known to be toxic and carcinogenic. Hope this answers the question. Have a nice day.
Ugo [173]3 years ago
4 0
The toxic gar expelled from the reaction between gasoline and oxygen in the vehicle's engine is both Carbon dioxide and monoxide
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The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
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Answer:

Approximately 2.46\; \rm mol.

Explanation:

Make use of the molar mass data (M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}) to calculate the number of moles of molecules in that 64.0\; \rm g of \rm C_2H_2:

\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}.

Make sure that the equation for this reaction is balanced.

Coefficient of \rm C_2H_2 in this equation: 2.

Coefficient of \rm H_2O in this equation: 2.

In other words, for every two moles of \rm C_2H_2 that this reaction consumes, two moles of \rm H_2O would be produced.

Equivalently, for every mole of \rm C_2H_2 that this reaction consumes, one mole of \rm H_2O would be produced.

Hence the ratio: \displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1.

Apply this ratio to find the number of moles of \rm H_2O that this reaction would have produced:

\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}.

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