Gọi 2a là số mol Na cần thiết

=>m_Na=23.2a=46a

2Na+2H₂O->2NaOH+H₂

2a......................2a........a...(mol)

Theo PTHH:m_NaOH=40.2a=80a

\(m_{H_2}\)=2a

Ta có:m_dd(sau)=500+46a-2a=500+44a

Theo gt:C_%dd=20%

=>\(\dfrac{80a}{500+44a}\).100%=20%

=>8000a=10000+880a=>7120a=10000

=>a=1,4(mol)

Vậy m_Na=46a=1,4.46=64,4(g)

m_HCl=30%.400=120(g)

=>n_HCl=120:36,5=3,3(mol)

m_Fe=0,1.56=5,6(g)

a)Fe+HCl->FeCl₂+H₂

0,1.......0,1.......0,1....0,1......(mol)

Ta có:\(\dfrac{n_{Fe}}{1}\)<\(\dfrac{n_{HCl}}{2}\)=>Fe hết,HCl dư

b)Theo PTHH:\(V_{H_2\left(đktc\right)}\)=0,1.22,4=2,24l

c)Theo PTHH:\(m_{FeCl_2}\)=0,1.127=12,7(g)

n_HCl(dư)=3,3-0,1=3,2(mol)

=>m_HCl(dư)=3,2.36,5=116,8(g)

Ta có:m_dd(sau)=5,6+400-0,1.2=405,4(g)

=>C_%HCl(dư)=\(\dfrac{116,8}{405,4}\).100%=28,8%

=>\(C_{\%FeCl_2}\)=\(\dfrac{12,7}{405,4}\).100%=3,13%

=50ml=0,05l

n_HCl=1.0,05=0,05(mol)

CuO+2HCl->CuCl₂+H₂O

0,025...0,05.....0,025..............(mol)

a)Theo PTHH:m_CuO(pư)=0,025.80=2(g)

b)Theo PTHH:\(n_{CuCl_2}\)=0,025(mol)

=>C_M(ddsau)=0,025:0,5=0,5M