In the formation of smog, nitrogen and oxygen gas react to form nitrogen dioxide

Question

In the formation of smog, nitrogen and oxygen gas react to form nitrogen dioxide. How many grams of NO2 will be produced when 2.0L of nitrogen at 840 mmHg and 24

Explanation / Answer

EXAMPLE n the formation of smog, nitrogen and oxygen gas react to form nitrogen dioxide: N2+2O2--->2NO2 How many grams of NO2 will be produced when 2.1 L of nitrogen at 820mmHg and 25 degrees celcius are completely reacted? ANSWER Step 1: Use the ideal gas law to calculate moles of N2: PV = n RT P = 820/760 = 1.08 atm V = 2.1 L T = 25+273 = 298 K (1.08)(2.1) = n (0.0821) (298) n = 0.0927 mol N2 2. Calculate moles of NO2 formed: 0.0927 mol N2 X (2 mol NO2/1 mol N2) = 0.185 mol NO2 3. Calculate mass of NO2: 0.185 mol NO2 X 46.0 g/mol = 8.53 grams NO2

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