\(N_{Fe}=0,25\cdot6\cdot10^{23}=1,5\cdot10^{23}\)

\(N_{Al}=1,5\cdot6\cdot10^{23}=9\cdot10^{23}\\ N_{H_2}=0,5\cdot6\cdot10^{23}=3\cdot10^{23}\\ N_{NaCl}=0,25\cdot6\cdot10^{23}=1,5\cdot10^{23}\\ n_{H_2O}=\dfrac{1,2\cdot10^{22}}{6\cdot10^{23}}=0,02\left(mol\right)\\ n_{Mg}=\dfrac{0,72\cdot10^{23}}{6\cdot10^{23}}=0,12\left(mol\right)\\ n_{CO_2}=\dfrac{3\cdot10^{22}}{6\cdot10^{23}}=0,05\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{80}{56\cdot2+16\cdot3}=0,5\left(mol\right)\\ n_{CaCO_3}=\dfrac{10}{40+12+16\cdot3}=0,1\left(mol\right)\\ n_{Na_2SO_4}=\dfrac{56,8}{23\cdot2+32+16\cdot4}=0,4\left(mol\right)\\ n_{K_2SO_4}=\dfrac{43,5}{39\cdot2+32+16\cdot4}=0,25\left(mol\right)\\ n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\\ n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)

\(m_{CuO}=0,25\cdot\left(64+16\right)=20\left(g\right)\\ m_{BaCl_2}=0,25\cdot\left(137+35,5\cdot2\right)=52\left(g\right)\\ m_{NaCl}=0,05\cdot\left(23+35,5\right)=2,925\left(g\right)\\ n_{H_2SO_4}=0,15\cdot\left(2+32+16\cdot4\right)=14,7\left(g\right)\)

thank you

1, Na₂O+H₂O---> 2NaOH

2, 2AL(OH)₃---> AL₂O₃+3H₂O

3, 2AL+6HCl--->2ALCL₃+3H₂

4, 2FE(OH)₃+3H₂SO₄---> FE₂(SO₄)₃+3H₂O

5, FE₂O₃+6HCL---> 2FECL₃+3H₂O

6,Zn+ H₂SO₄--> ZnSO₄+H₂

7, 3KCLO_3---> 3KCL+3O₂

8, CACO₃+2HCL---> CACL₂+CO₂+H₂O

9, P₂O₅+3H₂O---> 2H₃PO₄

10, AL₂O₃+3H₂SO₄---> AL₂(SO₄)₃+3H₂O

11, Mg+2HCL--->MGCL₂+H₂
12, 2NA+O₂--->2NA₂O

13 FECL₂+2NAOH--->FE(OH)₂+2NACL

14 4K+O₂---> 2K₂O

15, CAO+2HCL---> CACL₂+H₂O

\(n_{CO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ m_{CO_2}=44.0,6=26,4\left(g\right)\)

22,4 ở đâu thế bạn

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

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

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

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