For the following reaction, 20.4 grams of nitrogen monoxide are allowed to react with 5.10 grams of hydrogen gas . nitrogen monoxide(g) hydrogen(g) nitrogen(g) water(l) What is the maximum amount of nitrogen gas that can be formed

9.52g of N2

Explanation:

The equation of the reaction is given as;

2NO(g) + 2H2(g) → N2(g) + 2H2O(l)

From the stochiometry of the reaction;

2 mol   :    2 mol       :   1 mol     :   2 mol

60g     :      4g      :     28g   :  

( Mass = Number of moles * Molar mass)

Converting the masses given to moles;

Number of moles of NO = mass / molar mass = 20.4 / 30 = 0.68

Number of moles of H2 = mass / molar mass = 5.10 /  2 = 2.55

In this reaction, the limiting reactant is NO

60g of NO produces 28g of N2,

20.4g of NO would produce xg of N2

60 = 28

20.4 = X

X = (20.4 * 28) / 60

X = 9.52g

the answer is b & c i believe : )