\(n_{NO_2}=\dfrac{11,2}{22,4}=0,5mol\)
4NO2+2Ca(OH)2\(\rightarrow\)Ca(NO3)2+Ca(NO2)2+2H2O
\(n_{Ca\left(OH\right)_2}=\dfrac{1}{2}n_{NO_2}=0,25mol\)
\(m_{dd_{Ca\left(OH\right)_2}}=\dfrac{0,25.74.100}{7,4}=250g\)
\(n_{Ca\left(NO_3\right)_2}=n_{Ca\left(NO_2\right)_2}=\dfrac{1}{4}n_{NO_2}=\dfrac{0,5}{4}=0,125mol\)
\(C\%_{Ca\left(NO_3\right)_2}=\dfrac{0,125.164.100}{250}=8,2\%\)
\(C\%_{Ca\left(NO_2\right)_2}=\dfrac{0,125.132.100}{250}=6,6\%\)
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