a) Đặt: nMg=x(mol); nZnO=y(mol)

nH2SO4= 0,2(mol)

PTHH: Mg + H2SO4 -> MgSO4 + H2

x___________x____x_______x(mol)

ZnO + H2SO4 -> ZnSO4 + H2O

y____y______y(mol)

Ta có: 

\(\left\{{}\begin{matrix}24x+81y=12,9\\22,4x=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

mMg=0,2.24=4,8(g)

%mMg=(4,8/12,9).100=37,209%

=>%mZnO=62,791%

b) nH2SO4=x+y=0,3(mol)

=> \(C\%ddH2SO4=\dfrac{0,3.98}{120}.100=24,5\%\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

(mol)_____0,2____0,2______0,2____0,2__

\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)

\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)

\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)

\(a,n_{Fe}=\dfrac{22,4}{56}=0,4(mol)\\ n_{H_2SO_4}=\dfrac{24,5}{98}=0,25(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ LTL:\dfrac{0,4}{1}>\dfrac{0,25}{1}\Rightarrow Fe\text{ dư}\\ n_{Fe(dư)}=0,4-0,25=0,15(mol)\\ \Rightarrow m_{Fe(dư)}=0,15.56=8,4(g)\\ \)

\(b,m_{dư}=m_{Fe(dư)}=8,4(g)\\ c,n_{H_2}=0,25(mol)\\ \Rightarrow V_{H_2}=0,25.22,4=5,6(l)\\ d,n_{FeSO_4}=0,25(mol)\\\Rightarrow m_{FeSO_4}=0,25.152=38(g)\)

\(n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ a,n_{NaOH}=n_{H_2O}2.0,05=0,1\left(mol\right)\\ \Rightarrow m_{NaOH}=0,1.40=4\left(g\right)\\ b,C1:n_{Na_2SO_4}=n_{H_2SO_4}=0,05\left(mol\right)\\ m_{sp}=m_{Na_2SO_4}+m_{H_2O}=0,05.142+18.0,1=8,9\left(g\right)\\ C2:m_{sp}=m_{H_2SO_4}+m_{NaOH}=4,9+4=8,9\left(g\right)\)

\(a.\)

Theo tăng giảm khối lượng : 

\(n_{hh}=\dfrac{21.6}{96-60}=0.6\left(mol\right)\)

\(n_{CO_2}=n_{hh}=n_{H_2O}=0.6\left(mol\right)\)

Bảo toàn nguyên tố H : 

\(n_{H_2SO_4}=0.6\left(mol\right)\)

\(m_{dd_{H_2SO_4}}=\dfrac{0.6\cdot98}{24.5\%}=240\left(g\right)\)

\(b.\)

\(V_{CO_2}=0.6\cdot22.4=13.44\left(l\right)\)

2Al + 3H2SO4 => Al2(SO4)3 + 3H2

Fe + H2SO4 => FeSO4 + H2

m_{chất rắn không tan} = m_Cu = 7.68 (g)

Gọi x,y lần lượt là số mol của Al và Fe

Theo đề bài ta có:

171x + 152y = 81.7

1.5x + y = 14.56/22.4 = 0.65

Giải hệ phương trình ta được: x = 0.3, y = 0.2

m_Al = n.M = 0.3 x 27 = 8.1 (g)

m_Fe = n.M = 0.2 x 56 = 11.2 (g)

m_hh = 26.98 (g) ==> % m kim loại bạn tự tính nha

m_H2SO4 = 63.7g => m_{dd H2SO4} = 63.7x100/24.5 = 260 (g)

Dư 16% => Pứ = 100 - 16 = 84 %

m_{dd H2SO4 pứ} = 260x84/100 = 218.4 (g)

Gọi x,y lần lượt là số mol Fe, Al trong hh (x,y >0)

PTHH: Fe + H2SO4 -> FeSO4 + H2

x________x__________x_____x(mol)

2Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2

y____1,5y_________0,5y___1,5y(mol)

b) Ta có hpt:

\(\left\{{}\begin{matrix}56x+27x=11\\x+1,5y=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1=nFe\\y=0,2=nAl\end{matrix}\right.\)

=>mFe=0,1.56=5,6(g) ; mAl=0,2.27=5,4(g)

c) nH2SO4(tổng)=nH2=0,4(mol)

=> mH2SO4(tổng)=0,4.98=39,2(g)

=>mddH2SO4=(39,2.100)/24,5=160(g)

\(CuO+H_2SO_{4\left(24,5\%\right)}\rightarrow CuSO_4+H_2O\)

\(Cu+2H_2SO_{4đ}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)

\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(\Rightarrow n_{Cu}=0,05\left(mol\right)\)

\(\Rightarrow m_{CuO}=10-64.0,05=6,8\left(g\right)\)

\(\Rightarrow n_{CuO}=0,085\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(24,5\%\right)}=0,085\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(24,5\%\right)}=8,33\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4\left(24,5\%\right)}=34\left(g\right)\)