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How many moles of dinitrogen tetroxide are required to produce 5.6 grams of nitrogen dioxide?

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M(NO2) = 14.0 +2*16.0 = 46.0 g/mol
5.6 g NO2 *1 mol NO2/46.0 g NO2 ≈ 0.122 mol NO2

N2O4 -------> 2NO2
1 mol 2 mol
x mol 0.122 mol

x= 1*0.122/2 = 0.061 mol N2O4
User Abhishek Ghaskata
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