Let’s write down the solution according to the condition of the problem:
X l -? V = 7.5 liters.
1.2AgNO3 = 2Ag + 2NO2 + O2 – OBP, silver is obtained, nitric oxide is released (4);
M (NO2) = 46 g / mol;
M (O2) = 36 g / mol.
3. We make up the proportions:
1 mol of gas at normal level – 22.4 liters;
X mol (O2) – 7.5 liters. hence, X mol (O2) = 1 * 7.5 / 22.4 = 0.33 mol;
X mol (NO2) – 0.33 mol (О2);
-2 mol -1 mol hence, X mol (NO2) = 2 * 0.33 / 1 = 0.66 mol.
3. Let’s calculate the volume of the product:
V (NO2) = 0.66 * 22.4 = 14.78 liters.
Answer: during the reaction, nitrogen oxide with a volume of 14.78 liters was obtained.
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