nH2=V/22,4=3,36/22,4=0,15 (mol)
PT: Mg + 2HCl -> MgCl2 + H2
1........ 2............... .1.................1(mol)
0,15 <- 0,3 <- 0,15 <- 0,15 (mol)
mMg=n.M=0,15.24=3,6(g)
mHCl=n.M=0,3.36,5=10,95(g)
=>\(m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{10,95.100\%}{20\%}=54,75\left(g\right)\)