\(n_{HCl}=0,8.0,5=0,4(mol)\)
\(n_{H_{2}SO_{4}}=0,8.0,75=0,6(mol)\)
\(n_{H_{2}}=4,48:22,4=0,2(mol)\)
\(n_{Htrongaxitr}=0,4+2.0,6=1,6(mol)\)
Bảo toàn H: \(n_{H_{2}O}=\frac{n_{Htrongaxit}-2n_{H_{2}}}{2}=0,6(mol) \)
Bảo toàn khối lượng:
\(m=88,7+0,6.18+0,2.2−0,4.36,5−0,6.98=26,5(gam)\)