Câu 21: B

Câu 22: C

a) mN = 0,5 .14 = 7g.

mCl = 0,1 .35.5 = 3.55g

mO = 3.16 = 48g.

b) mN2 = 0,5 .28 = 14g.

mCl2 = 0,1 .71 = 7,1g

mO2 = 3.32 =96g

c) mFe = 0,1 .56 =5,6g mCu = 2,15.64 = 137,6g

mH2SO4 = 0,8.98 = 78,4g.

mCuSO4 = 0,5 .160 = 80g

1) \(n_{CO_2}=\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

\(V_{CO_2}=1,5.22,4=33,6\left(l\right)\)

2) \(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

3) \(V_{N_2}=0,025.22,4=0,56\left(l\right)\)

4) \(n_{SO_2}=\frac{32}{64}=0,5\left(mol\right)\)

\(V_{SO_2}=0,5.22,411,2\left(l\right)\)

5) \(n_{N_2}=\frac{2,8}{28}=0,1\left(mol\right)\)

\(n_{Cl_2}=\frac{7,1}{71}=0,1\left(mol\right)\)

\(V_{hh}=\left(0,1+0,1\right).22,4=4,48\left(l\right)\)

Chúc bạn học tốt

Bài này có ng làm r nên bài kia của bạn mk xóa đi nhé

a, \(n_{CO_2}=\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

-> \(V_{CO_2}=n_{CO_2}.22,4=1,5.22,4=33,6\left(l\right)\)

b, \(V_{O_2}=n_{O_2}.22,4=0,15.22,4=3,36\left(l\right)\)

c, \(V_{N_2}=n_{N_2}.22,4=0,025.22,4=0,56\left(l\right)\)

d, \(n_{SO_2}=\frac{m_{SO_2}}{M_{SO_2}}=\frac{32}{32+16.2}=0,5\left(mol\right)\)

-> \(V_{SO_2}=n_{SO_2}.22,4=0,5.22,4=11,2\left(l\right)\)

e, \(n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{2,8}{28}=0,1\left(mol\right)\)

-> \(V_{N_2}=n_{N_2}.22,4=0,1.22,4=2,24\left(l\right)\)

\(n_{Cl_2}=\frac{m_{Cl_2}}{M_{Cl_2}}=\frac{7,1}{71}=0,1\left(mol\right)\)

-> \(V_{Cl_2}=n_{Cl_2}.22,4=0,1.22,4=2,24\left(l\right)\)

=> \(V_{hh}=V_{N_2}+V_{Cl_2}=2,24+2,24=4,48\left(l\right)\)

a)

m_N = 0,5 . 14 = 7(g)

m_Cl = 0,1 . 35,5 = 3,55(g)

m_O = 3 . 16 = 48 (g)

b)

m_N2 = 0,5 . 28 = 14(g)

m_Cl2 = 0,1 . 71 = 7,1(g)

m_O2 = 3 . 32 = 96(g)

c)

m_Fe = 0,1 . 56 = 5,6(g)

m_Cu = 2,15 . 64 = 137,6(g)

m_H2SO4 = 0,8 . 98 = 78,4(g)

m_CuSO4 = 0,5 . 160 = 80(g)

Câu 3:

\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5mol\)

\(n_{Cu}=\dfrac{m}{M}=\dfrac{64}{64}=1mol\)

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2mol\)

Câu 4:

a)

m_N=n.M=0,5.14=7 gam

m_Cl=n.M=0,1.35,5=3,55 gam

m_O=n.M=3.16=48gam

b)

\(m_{N_2}=0,5.28=14gam\)

\(m_{Cl_2}=0,1.71=7,1gam\)

\(m_{O_2}=3.32=96gam\)

c)

m_Fe=0,1.56=5,6 gam

m_Cu=2,15.64=137,6 gam

\(m_{H_2SO_4}=0,8.98=78,4gam\)

\(m_{CuSO_4}=0,5.160=80gam\)

a, \(\overline{M}=\dfrac{0,1.44+0,2.28}{0,1+0,2}\approx33,33\left(g/mol\right)\)

b, \(\overline{M}=\dfrac{0,2.28+0,3.2}{0,2+0,3}=12,4\left(g/mol\right)\)

c, \(\overline{M}=\dfrac{0,1.28+0,2.30+0,2.44}{0,1+0,2+0,2}=35,2\left(g/mol\right)\)

d, \(\overline{M}=\dfrac{0,2.56+0,1.24+0,1.27}{0,2+0,1+0,1}=40,75\left(g/mol\right)\)

Hợp chất A

\(n_{Na}=\dfrac{4,6}{23}=0,2mol\\ n_{Cl}=\dfrac{7,1}{35,5}=0,2mol\)

     Na + Cl \(\rightarrow\) A

     0,2   0,2    0,2       ( mol ) \(\left\{Phần.này.bạn.không.hiểu.thì.hỏi.mình.nhé!\right\}\)

\(\Rightarrow\)   1      1      1       ( mol ) 

\(\Rightarrow CTHH:NaCl\)

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