Zero Formula 1.5 points A 20 -mm cube of copper metal is placed in 250 mL of $$ 12 \mathrm{Mnitricacid} \mathrm{at} 25^{\circ} \mathrm{C} $$ and the reaction below occurs: At a particular instant in time, nitrogen dioxide is being produced at the rate of $$ 23 imes 10^{-4} \mathrm{M} $$ min. At this same instant wheal ster erte twith hdrege ics reseen consumed in M/min? Do not write unit in answer. Report your answer with five places oast the darimal wins

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Zero Formula 1.5 points A 20 -mm cube of copper metal is placed in 250 mL of $$ 12 \mathrm{Mnitricacid} \mathrm{at} 25^{\circ} \mathrm{C} $$ and the reaction below occurs: At a particular instant in time, nitrogen dioxide is being produced at the rate of $$ 23 	imes 10^{-4} \mathrm{M} $$ min. At this same instant wheal ster erte twith hdrege ics reseen consumed in M/min? Do not write unit in answer. Report your answer with five places oast the darimal wins

UpStudy AI · Accepted Answer

The reaction between copper and nitric acid can be written as

$$
\text{Cu} + 4\text{HNO}_3 \longrightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}.
$$

The stoichiometric coefficients tell us that for every 2 moles of $$\text{NO}_2$$ produced, 4 moles of $$\text{HNO}_3$$ are consumed. This establishes the ratio

$$
\frac{4\,\text{mol HNO}_3}{2\,\text{mol NO}_2} = 2.
$$

Thus, the rate of consumption of nitric acid is twice the rate of production of nitrogen dioxide.

Given that nitrogen dioxide is produced at a rate of

$$
23 \times 10^{-4}\, \text{M/min},
$$

the rate of consumption of nitric acid is

$$
2 \times (23 \times 10^{-4})\, \text{M/min} = 46 \times 10^{-4}\, \text{M/min}.
$$

This can be expressed in decimal form as

$$
0.00460\, \text{M/min}.
$$

Thus, the consumption rate of nitric acid is

$$
0.00460.
$$
The nitric acid is consumed at a rate of 0.00460 M/min.

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