Bài 18: Mol

a: \(m=n\cdot M=0.5\cdot72=36\left(g\right)\)

\(a.m=n.M=0,5.72=36\left(g\right)\)

\(b.V=\dfrac{n}{22,4}=12,8:22,4=0,571\left(l\right)\)

\(\%Ca=\dfrac{1.40}{100}.100\%=40\%\\\%C=\dfrac{1.12}{100}.100\%=12\%\\ \%O=100\%-\left(40\%+12\%\right)=48\% \)

M_CaCO3 = 100 g/mol

%Ca = \(\dfrac{40}{100}.100\)= 40%

%C = \(\dfrac{12}{100}.100\)= 12%

%O₃= \(\dfrac{48}{100}.100\)=48%

Bài 4: 

a: \(n=\dfrac{28}{56}=0.5\left(mol\right)\)

b: \(n=\dfrac{32}{64}=0.5\left(mol\right)\)

Câu 2: C

Câu 3: C

2Fes+ 7/2O2 --->2SO2+FE2O3

\(4FeS+7O_2\underrightarrow{t^o}2Fe_2O_3+4SO_2\uparrow\)

4FeS + 7O2---> 4SO2 + 2Fe2O3

nếu bạn thấy đúng

Theo ĐLBTKL: m_S + m_O2 = m_SO2

=> m_O2 = 64-32 = 32(g)

2nguyên tử K

2

a. nMg=4.8/24=0,2(mol)

nCao=5.6/56=0,1(mol)

nCa=7.4/74=0,1(mol)

b. nC0₂= 4.48/22.4=0,2(mol)

\(V_{H_2}=0,6.24,79=14,874(l)\)

\(VH2=0,6.22,4=13,44\)\((l)\)

a) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)

b) \(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

c) \(n_{CO_2}=\dfrac{11}{44}=0,25\left(mol\right)\)

d) \(m_{O_2}=\dfrac{4,958.0,99}{0,082.\left(273+25\right)}=0,2\left(mol\right)\)

e) \(m_{CH_4}=\dfrac{12,359.0,99}{0,082\left(273+25\right)}=0,5\left(mol\right)\)

a: \(n=\dfrac{28}{56}=0.5\left(mol\right)\)

b: \(n=\dfrac{13.5}{27}=0.5\left(mol\right)\)