bài 1:

ta có phương trình:Zn+2HCl=>ZnCl2+h2

mZn=9.75(g)=>nZn=$\frac{9.75}{65}$=0.15(mol) 

a,nZn=0.15(mol)=>nHCl=0.15*2=0.3(mol)

=>mHCl=0.3*36.5=10.95(g)

b,nZnCl2=0.15(mol)=>mZnCl2=0.15*(65+35.5*2)=20.4(g)

c,3h2+Fe2O3=>2Fe+3H2O

nH2=0.15(mol)=>nFe2O3=0.15/3=0.05(mol)

=>mFe2O3=0.05*(56*2+16*3)=8(g)

nFe=$\frac{0.15*2}{3}$=0.1(mol)=>mFe=0.1*56=5.6(g) 

bài 2:

ta có phương trình:4Al+3O2=>2Al2O3

mAl=10.8(g)=>nAl=$\frac{10.8}{27}$=0.4(mol) 

a,ta có nAl2O3=$\frac{0.4*2}{4}$=0.2(mol) 

=>mAl2O3=0.2*(27*2+16*3)=20.4(g)

b,nO2=$\frac{0.4*3}{4}$=0.3(mol) =>vO2=0.3*22.4=6.72(lít)

c,ta có:2KMnO4=>K2MnO4+MnO2+O2

nO2=0.3(mol)=>nKMnO4=0.3*2=0.6(mol)

=>mKMnO4=0.6*(39+55+16*4)=94.8(g)

\(\begin{array}{l}
1)\\
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
nZn = \dfrac{{9,75}}{{65}} = 0,15\,mol\\
nHCl = 2nZn = 0,3\,mol\\
mHCl = 0,3 \times 36,5 = 10,95g\\
b)\\
nZnC{l_2} = nZn = 0,15\,mol\\
mZnC{l_2} = 136 \times 0,15 = 20,4g\\
c)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
n{H_2} = nZn = 0,15\,mol\\
nF{e_2}{O_3} = \dfrac{{0,15}}{3} = 0,05\,mol\\
mF{e_2}{O_3} = 0,05 \times 160 = 8g\\
nFe = \dfrac{{0,15 \times 2}}{3} = 0,1\,mol\\
mFe = 0,1 \times 56 = 5,6g\\
2)\\
a)\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
nAl = \dfrac{{10,8}}{{27}} = 0,4\,mol\\
nA{l_2}{O_3} = \dfrac{{0,4}}{2} = 0,2\,mol\\
mA{l_2}{O_3} = 0,2 \times 102 = 20,4g\\
b)\\
n{O_2} = \dfrac{{0,4 \times 3}}{4} = 0,3\,mol\\
V{O_2} = 0,3 \times 22,4 = 6,72l\\
c)\\
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
nKMn{O_4} = 0,3 \times 2 = 0,6\,mol\\
mKMn{O_4} = 0,6 \times 158 = 94,8g
\end{array}\)