\(n_{N_2}=\dfrac{0,224}{22,4}=0,01mol\\ 5Mg+12HNO_3\rightarrow5Mg\left(NO_3\right)_2+N_2+6H_2O\\ n_{Mg\left(NO_3\right)_2}=n_{Mg}=0,01.5=0,05mol\\ m_{Mg\left(NO_3\right)_2}=0,05.148=7,4g\\ n_{HNO_3}=0,01.12=0,12mol\\ m_{ddHNO_3}=\dfrac{0,12.63}{20\%}\cdot100\%=37,8g\\ m_{ddMg\left(NO_3\right)_2}=0,05.24+37,8-0,01.28=38g\\ C_{\%Mg\left(NO_3\right)_2}=\dfrac{7,4}{38}\cdot100\%=19,47\%\)