\(m_{CaO}=11,2\left(kg\right)=11200\left(g\right)\)

`=>` \(n_{CaO}=\dfrac{11200}{56}=200\left(mol\right)\)

PTHH: \(CaCO_3\xrightarrow[]{t^o}CaO+CO_2\)

             200<-------200

=> \(m_{đá.vôi}=\dfrac{200.100}{100\%-20\%}=25000\left(g\right)=25\left(kg\right)\)

\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)

            0,05<--0,1

=> \(V=\dfrac{0,05.22,4}{95\%}=\dfrac{112}{95}\left(l\right)\)

a) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂ 

           0,2<--0,4<------0,2<-----0,2

=> \(\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{ZnO}=21,1-13=8,1\left(g\right)\end{matrix}\right.\)

=> \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

b) PTHH: ZnO + 2HCl ---> ZnCl₂ + H₂O

                0,1---->0,2------->0,1

=> \(C\%_{HCl}=\dfrac{\left(0,2+0,4\right).36,5}{200}.100\%=10,95\%\)

c) \(m_{muối}=m_{ZnCl_2}=\left(0,1+0,2\right).136=40,8\left(g\right)\)

a) \(n_{CuSO_4}=\dfrac{32.10\%}{160}=0,02\left(mol\right)\)

PTHH: Zn + CuSO₄ ---> ZnSO₄ + Cu (Phản ứng thế)

            0,02<--0,02----->0,02------>0,02

b) m_{Zn (pư)} = 0,02.65 = 1,3 (g)

c) m_dd = 32 + 0,02.65 - 0,02.64 = 32,02 (g)

=> \(C\%_{ZnSO_4}=\dfrac{0,02.161}{32,02}.100\%=10,056\%\)

nCO₂= VCO₂ / 22,4=6,72 / 22,4=0,3(mol)

a.CO₂+Ba(OH)₂->BaCO₃\(\downarrow\)+H₂O

__0,3__0,3______0,3

b.CMBa(OH)₂=n / V=0,3/ 0,6=0,5 (mol/l)

c.mBaCO₃=n.M=0,3 . 197=59,1(g)

n_khí=V_k/22,4=10,08/22,4=0,45 (mol)

a.PT:Fe + HCl ->FeCl₂ +H₂\(\uparrow\)

___0,45__0,9_________0,45

b.m_Fe=n_Fe . M_Fe= 0,45 . 56= 25,2 (g)

c.CM_HCl= n_HCl / V_HCl= 0,9 / 0,15 =6 (mol/l)