Question

2. If the percent yield for the following reaction is 75.0%, and 25.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced? 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

Answer 1

2)

Molar mass of NO2,

MM = 1*MM(N) + 2*MM(O)

= 1*14.01 + 2*16.0

= 46.01 g/mol

mass of NO2 = 25 g

mol of NO2 = (mass)/(molar mass)

= 25/46.01

= 0.5434 mol

According to balanced equation

mol of HNO3 formed = (2/3)* moles of NO2

= (2/3)*0.5434

= 0.3622 mol

Molar mass of HNO3,

MM = 1*MM(H) + 1*MM(N) + 3*MM(O)

= 1*1.008 + 1*14.01 + 3*16.0

= 63.018 g/mol

mass of HNO3 = number of mol * molar mass

= 0.3622*63.02

= 22.83 g

% yield = actual mass*100/theoretical mass

75.0= actual mass*100/22.83

actual mass=17.12 g

Answer: 17.1 g

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