bài này là dạng cơ bản , chỉ cần áp dụng công thức V = n*22,4 

nCO2= 0.44/44=0.01mol

VCO2= 0.224 l

nH2=1/2=0.5 mol

VH2= 0.5*22.4=11.2l

Thể tích hỗn hợp khí:0.224+11.2=11.424l

\(a.m_{Mg}=0,1.24=2,4\left(g\right)\\ m_{Ca}=0,2.40=8\left(g\right)\\ b.n_{hh}=\dfrac{2,8}{28}+\dfrac{13,2}{44}=0,4\left(mol\right)\\ \Rightarrow V_{hh}=0,4.22,4=8.96\left(l\right)\)

a) n_Fe= \(\frac{5,6}{56}\)= 0,1 mol

n_Cu= \(\frac{64}{64}\)= 1mol

n_Al= \(\frac{27}{27}\)= 1 mol

b)

n_CO2= \(\frac{44}{12+16.2}\)= 1 mol

n_H2= \(\frac{4}{1.2}\)= 2 mol

=> n_hh= 1+2= 3 mol

V_hh= 3.22,4= 67,2 l

a) Số mol Fe trong 5,6 g Fe:

n_Fe=\(\frac{m_{Fe}}{M_{Fe}}=\frac{5,6}{56}=0,1\left(mol\right)\)

Số mol Cu có trong 64 g Cu:

n_Cu=\(\frac{m_{Cu}}{M_{Cu}}=\frac{64}{64}=1\left(mol\right)\)

Số mol Al có trong 27 g Al:

n_Al= \(\frac{m_{Al}}{M_{Al}}=\frac{27}{27}=1\left(mol\right)\)

a) n_Fe= \(\frac{m_{Fe}}{M_{Fe}}=\frac{5,6}{56}=0,1\left(mol\right)\)

n_Cu=\(\frac{m_{Cu}}{M_{Cu}}=\frac{64}{64}=1\left(mol\right)\)

n_Al= \(\frac{m_{Al}}{M_{Al}}=\frac{27}{27}=1\left(mol\right)\)

b) \(n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{44}{44}=1\left(mol\right)\)

\(n_{H_2}=\frac{m_{H_2}}{M_{H_2}}=\frac{4}{2}=2\left(mol\right)\)

a) n_Fe = 5,6/56 = 0,1 mol

n_Cu = 64/64 = 1 mol

n_Al = 27/27 = 1 mol

b) n_CO2 = 44/44 = 1 mol

=> V_CO2 = 1.22,4 = 22,4 l

n_H2 = 4/2 = 2 mol

=> V_H2 = 2.22,4 = 44,8 l

a) n_Fe = \(\frac{5.6}{56}=0.1\)(mol)

n_Cu = \(\frac{64}{64}=1\) (mol)

n_Al = \(\frac{27}{27}=1\) (mol)

b) n_CO2 = \(\frac{44}{44}=1\)(mol)

n_H2 = \(\frac{4}{2}=2\) (mol)

\(\Rightarrow\)n_hh = 1 + 2 = 3 (mol)

\(\Rightarrow\)V_hh = 3 * 22.4 = 67.2 (l)

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

\(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)

b) 

\(m_{CO_2}=44.0,5=22\left(g\right)\)

\(m_{H_2}=1,5.2=3\left(g\right)\)

\(m_{N_2}=2.28=56\left(g\right)\)

\(m_{CuO}=3.80=240\left(g\right)\)

c) \(n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\)

\(n_{H_2}=\dfrac{4,8}{2}=2,4\left(mol\right)\)

\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)

=> n_hh = 0,2 + 2,4 + 0,1 = 2,7 (mol)

=> V_hh = 2,7.22,4 = 60,48(l)

\(a.n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\\ n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)

\(b.m_{CO_2}=0,5.44=22\left(g\right)\\ m_{H_2}=1,5.2=3\left(g\right)\\ m_{N_2}=2.28=56\left(g\right)\\ m_{CuO}=3.80=240\left(g\right)\)

\(c.n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{4,8}{2}=2,4\left(mol\right)\\ n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\\\Rightarrow n_{hh}=0,2+2,4+0,1=2,7\left(mol\right)\\ \Rightarrow V_{hh}=2,7.22,4=60,48\left(l\right)\)

\(a,n_{Mg}=\dfrac{2,4}{24}=0,1(mol); n_{Zn}=\dfrac{32,5}{65}=0,5(mol)\\ n_{Al}=\dfrac{2,7}{27}=0,1(mol); n_{Cu}=\dfrac{19,2}{64}=0,3(mol)\)

\(b,m_{CO_2}=0,5.44=22(g);m_{H_2}=1,5.2=3(g)\\ m_{N_2}=2.28=56(g);m_{CuO}=3.80=240(g)\)

\(c,n_{hh}=n_{Cl_2}+n_{H_2}+n_{O_2}=\dfrac{14,2}{71}+\dfrac{4,8}{2}+\dfrac{3,2}{32}=0,2+2,4+0,1=2,7(mol)\\ V_{hh}=2,7.22,4=60,48(l)\)

a)

b) VCO2 = 22,4 .0,175 = 3,92l.

VH2 = 22,4 .1,25 = 28l.

VN2 = 22,4.3 = 67,2l.

c) Số mol của hỗn hợp khí bằng tổng số mol của từng khí.

nhh = nCO2 + nH2 + nN2 = 0,01 + 0,02 + 0,02 = 0,05 mol

Vhh khí = (0,01 + 0,02 + 0,02) . 22,4 = 1,12l.

Bài 1:

a) \(V_{khí}=\left(0,2+0,5+0,35\right)\cdot22,4=23,52\left(l\right)\)

b) \(m_{khí}=0,2\cdot64+0,5\cdot28+0,35\cdot28=36,6\left(g\right)\)

Bài 2:

Sửa đề: Tỉ lệ \(Mg:C:O=2:1:4\)

Ta có: \(n_{Mg}:n_C:n_O=\dfrac{2}{24}:\dfrac{1}{12}:\dfrac{4}{16}=1:1:3\)

\(\Rightarrow\) CTHH là MgCO₃