a: \(n_{SO_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right);n_{NaOH}=0.1\cdot2=0.2\left(mol\right)\)
\(T=\dfrac{0.2}{0.25}=0.8< 1\)
=>Chỉ có muối \(NaHSO_3\)
\(SO_2+NaOH\rightarrow NaHSO_3\)
0,2 0,2 0,2
=>SO2 dư ra 0,05mol
\(m=0.2\cdot104+0.05\cdot64=24\left(g\right)\)
b: \(n_{NaOH}=4\cdot0.1=0.4\left(mol\right)\)
\(n_{SO_2}=0.25\left(mol\right)\)
\(T=\dfrac{0.4}{0.25}=1.6\)
=>1<T<2
=>Tạo ra 2 muối \(NaHSO_3;Na_2SO_3\)
Đặt \(n_{SO_2}=x\left(mol\right);n_{NaOH}=y\left(mol\right)\)
\(SO_2+NaOH\rightarrow NaHSO_3\)
x x x
\(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)
y 2y 2y
Do đó, ta có hệ phương trình:
\(\left\{{}\begin{matrix}x+y=0.25\\x+2y=0.4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0.1\\y=0.15\end{matrix}\right.\)
\(m=0.1\cdot104+0.15\cdot126=29.3\left(g\right)\)
c: \(n_{NaOH}=6\cdot0.1=0.6\left(mol\right);n_{SO_2}=0.25\left(mol\right)\)
\(T=\dfrac{0.6}{0.25}=2.4>2\)
=>Chỉ có muối Na2SO3
\(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)
0,25 0,5 0,25
=>NaOH dư ra 0,1mol
\(m=0.25\cdot126+0.1\cdot40=35.5\left(g\right)\)