Câu 24:

\(V_{H_2}=0,125.22,4=2,8\left(l\right)\)

→ Đáp án: D

Câu 25:

\(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Theo PT: \(n_{H_2}=n_{CuO}=0,5\left(mol\right)\) \(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)

→ Đáp án: B

Câu 26:

\(m_{ZnO}=0,5.81=40,5\left(g\right)\)

→ Đáp án: A

Câu 27: B

Câu 28: B

N phân tử = 1 mol phân tử

\(\Rightarrow n_{O2}=1mol;n_{N_2}=2mol;n_{CO_2}=1,5mol\)

\(\Rightarrow m_{hh}=1.32+2.28+1,5.44=154g\)

b. \(m_{hh}=0,1.56+0,2.64+0,3.65+0,25.27=44,65g\)

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

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)

\(n_{HCl}=\dfrac{6,72}{22,4}=0,3mol\)

\(n_{CO_2}=\dfrac{0,56}{22,4}=0,025mol\)

\(\Rightarrow m_{hh}=0,1.32+0,05.2+0,3.36,5+0,025.44=15,35g\)

a)

$n_{Al_2(SO_4)_3} = \dfrac{75,24}{27.2 + 32.3 + 16.12} = 0,22(mol)$

b)

$n_{O_2} = \dfrac{15,68}{22,4} = 0,7(mol)$
c)

$n_{H_2SO_4} = \dfrac{13,2.10^{23}}{6.10^{23}} = 2,2(mol)$

d)

$n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$n_{Al} = \dfrac{3,24}{27} = 0,12(mol)$
$n_{X} = 0,2 + 0,12 = 0,32(mol)$

e)

$n_{O_2} = \dfrac{8,94}{22,4} = 0,4(mol)$
$n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$n_Y = 0,4 + 0,1 = 0,5(mol)$

a) nAl2(SO4)3= mAl2(SO4)3/M(Al2(SO4)3)= 75,24/342=0,22(mol)

b) nO2=V(O2,đktc)/22,4=15,68/22,4=0,7(mol)

c) nH2SO4=N/6.10²³= (13,2.10²³)/(6.10²³)= 2,2(mol)

d) nX có:

Số mol Fe: nFe= mFe/M(Fe)=11,2/56=0,2(mol)

Số mol Al: nAl=mAl/M(Al)=3,24/27=0,12(mol)

e) nY có:

Số mol O2: nO2=V(O2,đktc)/22,4=8,94/22,4=447/1120(mol)

Số mol H2: nH2=V(H2,đktc)/22,4=2,24/22,4=0,1(mol

a, m_CaO = 0,5.56 = 28 (g)

b, \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

c, \(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

d, \(V_{hhk}=0,2.22,4+0,3.22,4=11,2\left(l\right)\)

e, \(\%m_{Cu}=\dfrac{64}{64+32+16.4}.100\%=40\%\)

Bạn tham khảo nhé!

a) mCaO=nCaO.M(CaO)=0,5.56=28(g)

b) nCO2=V(CO2,dktc)=6,72/22.4=0,3(mol)

c) nH2SO4=mH2SO4/M(H2SO4)=24,5/98=0,25(mol)

d) V(hh H2,NH3)=(0,3+0,2).22,4=11,2(l)

e) %mCu/CuSO4=(64/160).100=40%

Chúc em học tốt!

Ta có: \(n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)\)

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

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(\Rightarrow\overline{M}_B=\dfrac{m_{CO_2}+m_{O_2}+m_{H_2}}{n_{CO_2}+n_{O_2}+n_{H_2}}=\dfrac{4,4+0,1.32+0,3.2}{0,1+0,1+0,3}=16,4\left(g/mol\right)\)

\(\Rightarrow d_{B/H_2}=\dfrac{\overline{M}_B}{M_{H_2}}=\dfrac{16,4}{2}=8,2\)

\(a,n_{Fe(OH)_3}=\dfrac{5,35}{107}=0,05(mol)\\ b,n_{NH_3}=\dfrac{0,56}{22,4}=0,025(mol)\\ c,n_{CO_2}=\dfrac{11}{44}=0,25(kmol)\\ d,n_{H_2}=\dfrac{0,448}{22,4}=0,02(mol)\)

a) \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

b) \(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\)

c) \(n_{N_2}=\dfrac{5,6}{28}=0,2\left(mol\right)\)

a,hoàn thành PTHH sau và cho biết tỉ lệ số nguyên tử (phân tử) các chất trong phản ứng hóa học sau

- 2Mg + O2 --to->  2MgO

=>2:1:2

- 4Fe +3 O2 -to--> 2Fe2O3

4:3:2

b, -Tính khối lượng mol của Al2O3

=>MAl2O3=102đvC

- Số mol của 20,4g Al2O3, 2,24 lít   CO2 (đktc)

=>n Al2O3=2mol

n CO2=0,1 mol

\(a.n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\\ n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ n_{H_3PO_4}=\dfrac{11,76}{98}=0,12\left(mol\right)\\ n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\\ b.n_{C_2H_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ n_{N_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

a) sửa CaO

`n_(CaO)=(14/56)=0,25`mol

b)

`n_(CO_2)= 66/44=1,5` mol

c)

`n_(H_2)= (3,36)/22,4=0,15 mol`

d)

`n_(SO_3)= (22,4)/22,4=0,1 mol`

e)

`n_(Mg)=(3,55.10^23)/(6,02.10^23)=0,589mol.`

f)

tương tự e)