\(a)7437ml=7,437l\\ n_{N_2}=\dfrac{7,437}{24,79}=0,3mol\\ m_{N_2}=0,3.28=8,4g\\ b)n_{Cl_2}=\dfrac{2,479}{24,79}=0,1mol\\ m_{Cl_2}=0,1.71=7,1g\)

b, 14,9 g

\(n_{H_2}=\dfrac{2,479}{24,79}=0,1mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{Mg}=n_{H_2}=0,1mol\\ m_{Mg}=0,1.24=2,4g\)

a, \(n_{CO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(MgO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)

\(MgCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)

Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,1.84}{10,4}.100\%\approx80,77\%\\\%m_{MgO}\approx19,23\%\end{matrix}\right.\)

b, \(n_{MgO}=\dfrac{10,4-0,1.84}{40}=0,05\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{MgO}+2n_{MgCO_3}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

a) \(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{2.479}{22,4}\approx0,11\left(mol\right)\)

b) \(m_{CO_2}=n.M=0,11=4,84\left(g\right)\)

c) Số phân tử CO₂ là : \(0,11.6.10^{23}=0,66.10^{23}\)

a) \(n_{CO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

b) \(m_{CO_2}=0,1.44=4,4\left(g\right)\)

c) Số phân tử CO₂ = 0,1.6.10²³ = 0,6.10²³

Câu 6:

1. \(n_{O_2}=\dfrac{2,470}{24,79}=0,1\left(mol\right)\)

PTHH:

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

2/15    0,1              1/15

\(m_{Al}=\dfrac{2}{15}.27=3,6\left(g\right)\)

2. Gọi m KCl cần thêm là x 

Ta có:

\(15\%=\dfrac{\dfrac{10x.10}{100}+\dfrac{300.25}{100}}{10x+300}\)

\(\Rightarrow x=60\)

Vậy \(m_{ddKCl}=\dfrac{60.100}{10}=600\left(g\right)\)

\(n_{Cl_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PTHH: \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

             0,04<-----------------------------------------------0,1

\(\Rightarrow n_{KMnO_4\left(c\text{ần}.d\text{ùng}\right)}=\dfrac{0,04}{80\%}=0,05\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,05.158=7,9\left(g\right)\)

PT: \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

Ta có: \(n_{Cl_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

Theo PT: \(n_{KMnO_4\left(LT\right)}=\dfrac{2}{5}n_{Cl_2}=0,04\left(mol\right)\)

Mà: H% = 80% \(\Rightarrow n_{KMnO_4\left(TT\right)}=\dfrac{0,04}{80\%}=0,05\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=0,05.158=7,9\left(g\right)\)

a.

\(m_{Al}=0.5\cdot27=13.5\left(g\right)\)

\(m_{CO_2}=\dfrac{6.72}{22.4}\cdot44=13.2\left(g\right)\)

\(m_{N_2}=\dfrac{5.6}{22.4}\cdot28=7\left(g\right)\)

\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)

b.

\(m_{hh}=\dfrac{3.36}{22.4}\cdot2+\dfrac{5.6}{22.4}\cdot28+0.2\cdot44=16.1\left(g\right)\)