`a.`
`n_{Al}=m/M=(0,54)/27=0,02(mol)`
`b.`
`n_{NaCl}=m/M=(11,7)/(58,5)=0,2(mol)`
`c.`
`n_{CuSO_4 .5H_2O}=m/M=125/250=0,5(mol)`
`d.`
`n_{Fe}=N/(6.10^23)=(18,066.10^22)/(6.10^23)=0,3011(mol)`
`e.`
`n_{O_2}=V/(22,4)=(1,12)/(22,4)=0,05(mol)`
`f.`
`200(ml)=0,2(l)`
`n_{H_2SO_4}=C_M .V_{dd}=0,6.0,2=0,12(mol)`
`\text{___________________________________}`
Đáp án:
`a.n_{Al}=0,02(mol)`
`b.n_{NaCl}=0,2(mol)`
`c.n_{CuSO_4 .5H_2O}=0,5(mol)`
`d.n_{Fe}=0,3011(mol)`
`e.n_{O_2}=0,05(mol)`
`f.n_{H_2SO_4}=0,12(mol)`