A mixture of and contains nitrogen. The mass ratio of and in the mixture is: — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Let the mass of $NH_4NO_3$ ($M=80$) be $x$ and $(NH_4)_2HPO_4$ ($M=132$) be $y$. <br>Nitrogen in $NH_4NO_3 = (28/80)x = 0.35x$. <br>Nitrogen in $(NH_4)_2HPO_4 = (28/132)y = 0.212y$. <br>Total Nitrogen = $0.304(x+y)$. <br>$0.35x + 0.212y = 0.304x + 0.304y \Rightarrow 0.046x = 0.092y \Rightarrow x/y = 2/1$.