Đáp án:
`1, `
`a, m=30 gam; V=6,72 lít`
`b, 19,96%`
`2,`
`a, a=16,25%; m=21,4 gam`
`b, 7,33%`
Giải thích các bước giải:
Bài 1:
`CaCO_3 + 2HCl -> CaCl_2 + CO_2 + H_2O`
`a,`
`m_{HCl}=150.14,6%=21,9(gam)`
`-> n_{HCl}=(21,9)/(36,5)=0,6(mol)`
Theo PTHH:
`n_{CaCO_3}=1/2 .n_{HCl}=1/2 .0,6=0,3(mol)`
`=> m_{CaCO_3}=0,3.100=30(gam)`
`n_{CO_2}=1/2 .n_{HCl}=1/2 .0,6=0,3(mol)`
`=> V_{CO_2}=0,3.22,4=6,72(lít)`
`b,`
`m_{dd sau pứ}=30 + 150 - 0,3.44=166,8(gam)`
Theo PTHH:
`n_{CaCl_2}=n_{1/2 .n_{HCl}=1/2 .0,6=0,3(mol)`
`-> m_{CaCl_2}=0,3.111=33,3(gam)`
`=> C%_{CaCl_2}=(33,3)/(166,8).100%=19,96%`
Bài 2:
`3NaOH + FeCl_3 -> Fe(OH)_3 + 3NaCl`
`a,`
`m_{NaOH}=300.8%=24(gam)`
`-> n_{NaOH}=(24)/(40)=0,6(mol)`
Theo PTHH:
`n_{FeCl_3}=1/3 .n_{NaOH}=1/3 .0,6=0,2(mol)`
`-> m_{FeCl_3}=162,5 . 0,2=32,5(gam)`
`=> a=(32,5)/(200).100%=16,25%`
Theo PTHH:
`n_{Fe(OH)_3}=1/3 .n_{NaOH}=1/3 .0,6=0,2(mol)`
`-> m_{Fe(OH)_3}=0,2.107=21,4(gam)`
`b,`
`m_{dd sau pứ}=300 + 200 - 21,4=478,6(gam)`
`n_{NaCl}=n_{NaOH}=0,6(mol)`
`-> m_{NaCl}=0,6.58,5=35,1(gam)`
`=> C%_{NaCl}=(35,1)/(478,6).100%=7,33%`
