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aksik [14]
3 years ago
13

How many grams of NO will be produced from 60.0g of NO2 reacted with excess water in the following chemical reaction?

Chemistry
1 answer:
Lynna [10]3 years ago
6 0

Answer:

Mass = 19.78 g

Explanation:

Given data:

Mass of NO produced = ?

Mass of NO₂ reacted = 60.0 g

Solution:

Chemical equation:

3NO₂(g) + H₂O(l)   →  2HNO₃(g) + NO(g)

Number of moles of NO₂:

Number of moles = mass/molar mass

Number of moles = 60.0 g/ 46 g/mol

Number of moles = 1.3 mol

Now we will compare the moles of NO₂ with NO.

                      NO₂         :           NO

                        3            :            1

                      1.3            :            1/3×1.3 = 0.43 mol

Mass of NO:

Mass = number of moles × molar mass

Mass = 0.43 mol × 46 g/mol

Mass = 19.78 g

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Approximately 0.291\; \rm M (rounded to two significant figures.)

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\rm HCl reacts with \rm NaOH at a one-to-one ratio:

\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l).

Coefficient ratio:

\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1.

In other words, one mole of \rm NaOH would neutralize exactly one mole of \rm HCl. In this titration, 0.291\; \rm mol of \rm NaOH\! was required. Therefore, the same amount of \rm HC should be present in the original solution:

\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}.

Calculate the concentration of the original \rm HCl solution:

\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M.

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