a) Gọi \(\left\{{}\begin{matrix}n_{NaNO_3}=a\left(mol\right)\\n_{Cu\left(NO_3\right)_2}=b\left(mol\right)\end{matrix}\right.\)
=> 85a + 188b = 4,43 (1)
PTHH: \(2NaNO_3\xrightarrow[]{t^o}2NaNO_2+O_2\)
a-------------------------->0,5a
\(2Cu\left(NO_3\right)_2\xrightarrow[]{t^o}2CuO+4NO_2+O_2\)
b------------------------>2b------->0,5b
MA = 19,5.2 = 39 (g/mol)
Áp dụng sơ đồ đường chéo, ta có:
\(\dfrac{n_{NO_2}}{n_{O_2}}=\dfrac{39-32}{46-39}=1\)
=> \(n_{NO_2}=n_{O_2}\)
=> 2b = 0,5a + 0,5b
=> 0,5a - 1,5b = 0 (2)
Từ (1), (2) => a = 0,03; b = 0,01
=> \(V_A=\left(0,5.0,03+0,01.2+0,01.0,5\right).22,4=0,896\left(l\right)\)
b) \(\left\{{}\begin{matrix}m_{NaNO_3}=0,03.85=2,55\left(g\right)\\m_{Cu\left(NO_3\right)_2}=0,01.188=1,88\left(g\right)\end{matrix}\right.\)
