Question

A mixture of NH4NO3 and (NH4)2HPO4 contain 30.40 % mass percent of nitrogen .What is the mass ratio of the two components in the mixture ??

Answer 1

NH4NO3: molar mass = N (14.01) + 4H (4 x 1.008) + N (14.01) + 3O (3 x 16.00) = 80.05 g/mole 
%N = (2N / 80.05) x 100 = ((2 x 14.01) / 80.05) x 100 = 35.00% 

(NH4)2HPO4: molar mass = 2N (2 x 14.01) + 8H (8 x 1.008) + H (1.008) + P (30.97) + 4O (4 x 16.00) = 132.06 g/mole. 
%N = (2N / 132.06) x 100 = ((2 x 14.01) / 132.06)) x 100 = 21.22% 

Let's say we have 100 grams of the mixture. 
Let x = grams of NH4NO3 
Let y = grams of (NH4)2HPO4 

x + y = 100 g 
x = 100 - y 

0.3500x + 0.2122y = 0.3047(100g) . . .substitute (100 - y) for x 
(0.3500)(100 - y) + 0.2122y = 30.47 
35.00 - 0.3500y + 0.2122y = 30.47 
4.53 = 0.1378y 
y = 32.9 g 
x = 100 - y = 100 - 32.9 = 67.1 g

Answer 2

Answer:

this is your answer mate,.....,