\(n_X=\dfrac{v}{22,4}=\dfrac{31,36}{22,4}=1,4mol\)
- Dựa vào tỉ lệ mol ta gọi: \(n_{O_2}=2x;n_{N_2}=3x;n_{CO_2}=2x\)
-Ta có: 2x+3x+2x=1,4\(\rightarrow\)7x=1,4\(\rightarrow\)x=0,2mol
\(\rightarrow\)\(n_{O_2}=2x=0,4mol;n_{N_2}=3x=0,6mol;n_{CO_2}=2x=0,4mol\)
\(\rightarrow\)\(m_X=0,4.32+0,6.28+0,4.44=47,2gam\)