mHCl=\(\dfrac{C\%.m_{ddHCl}}{100\%}=\dfrac{10\%.219}{100\%}=21,9\left(g\right)\)
\(n_{HCl}=\dfrac{m}{M}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
nMg=m/M=6/24=0,25(mol)
Pt: Mg+2HCl-> MgCl2+H2
1......2............1...........1 (mol)
0,25....0,5.........0,25.....0,25 (mol)
Vậy chất dư sau phản ứng là HCl
số mol HCl dư là 0,6 - 0,5 =0,1(mol)
mHCl dư =ndư.M=0,1.36,5=3,65(g)
b) VH2=n.22,4=0,25.22,4=5,6(g)
c) md d sau phản ứng=mMg+mHCl-mH2=6+219-(0,25.2)=224,5(g)
=> \(C\%_{MgCl_2}=\dfrac{m_{MgCl_2}.100\%}{m_{ddsauphanung}}=\dfrac{n.M.100\%}{224,5}=\dfrac{0,25.95.100\%}{224,5}=10,57\left(\%\right)\)