Question

how many grams of nitrogen oxide two needed to win 20dm3 nitrogen oxide four.
2NO+02--->2NO2

Answer 1

Answer on Question #64984 - Chemistry – Other

Task:

How many grams of nitrogen oxide two (NO) needed to win $20\mathrm{dm}^3$ nitrogen oxide four $(\mathrm{NO}_2)$.

$2\mathrm{NO} + \mathrm{O}_2 \longrightarrow 2\mathrm{NO}_2$

Solution:

The equation of the chemical reaction:

$$2 NO + O _ {2} = 2 NO _ {2}$$

By the equation:

$$\nu (N O) = \nu (N O _ {2});$$ $$\frac {m (N O)}{M (N O)} = \frac {V (N O _ {2})}{V m};$$

The molar volume of an ideal gas $(V_{\mathrm{m}})$ at 1 atmosphere of pressure is $22.414~\mathrm{dm}^3/\mathrm{mol}$ at $0^{\circ}\mathrm{C}$. The molar mass of nitrogen oxide (II): M(NO) = 30.01 g*mol$^{-1}$.

Then,

$$m (N O) = \frac {V (N O _ {2})}{V m} \times M (N O) = \frac {2 0 \times 3 0 . 0 1}{2 2 . 4 1 4} = 2 6. 7 7 7 9 \approx 2 6. 7 8 (g)$$

Answer: $m(NO) = 26.78$ g.

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