\(n_{H_2}=0,4\left(mol\right)\)

\(Zn+H_2SO_4-->ZnSO_4+H_2\uparrow\)

x.........x...............................x............x

\(2Al+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2\uparrow\)

y.........1,5y.......................0,5y.................1,5y

\(\left\{{}\begin{matrix}65x+27y=11,9\\x+1,5y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\%Zn=\dfrac{0,1.65}{11,9}.100\%\approx54,62\%\)

\(\%Al=100\%-54,62\%=45,38\%\)

b) nH2=nH2SO4=0,4(mol)

\(V_{ddH_2SO_4}=\dfrac{0,4}{0,5}=0,8\left(l\right)\)

theo BTKL

\(m_{FeCl_3}=m_{Fe}+m_{HCl}=11,2+8,96=20,16\left(g\right)\)

1) \(2Cu\left(NO_3\right)_2-t^0->2CuO+4NO_2\uparrow+O_2\uparrow\)

NO2 nha bn

\(\)2) \(Cu+2H_2SO_4\left(đ\right)-t^0->CuSO_4+SO_2\uparrow+2H_2O\)

3) \(4FeS_2+11O_2-t^0->2Fe_2O_3+8SO_2\uparrow\)

nếu Nguyễn Khắc Vinh thấy dễ thì làm đi

\(n_{Zn}=0,1\left(mol\right)\)

\(n_{HCl}=0,4\left(mol\right)\)

\(Zn+2HCl-->ZnCl_2+H_2\uparrow\)

\(\dfrac{0,1}{1}< \dfrac{0,4}{2}\) => Zn hết HCl dư

a) \(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

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

\(m_{KClO_3}pu=\dfrac{24,5.80\%}{100\%}=19,6g\)

theo BTKL

\(m_{O_2}=m_{KClO_3pu}-m_{KCl}=19,6-13,45=6,15\left(g\right)\)

bài 1:

VH2O=800ml=0,8l

\(n_{NaCl}=\dfrac{11,7}{58,5}=0,2\left(mol\right)\)

\(C_{Mdd}=\dfrac{0,2}{0,8}=0,25\left(M\right)\)

bai 2

\(C\%_{FeSO_4}=\dfrac{18,3}{18,3+181,7}.100\%=9,15\%\)

\(200=2^3.5^2\)

=>200 co so uoc la : (3+1).(2+1)=12 (uoc)

\(n_{H_2SO_4}=\dfrac{250.20\%}{98.1005}\approx0,51\left(mol\right)\)

\(m_{H_2SO_4}=0,51.98=49,98\left(g\right)\)

\(m_{dd\left(bd\right)}=\dfrac{49,98.100\%}{74\%}=67,54\left(g\right)\)

\(V_{dd\left(bd\right)}=\dfrac{67,54}{1,664}\approx40,58\left(ml\right)\)

\(n_{HCl}=1.2=2\left(mol\right)\)

\(m_{HCl}=2.36,5=73\left(g\right)\)

\(m_{dd\left(bd\right)}=\dfrac{73.100\%}{38\%}=192,1\left(g\right)\)

\(V_{dd\left(bd\right)}=\dfrac{192,1}{1,19}=161,43\left(ml\right)\)