Trộn lẫn 200g dung dịch KOH 7% với 100 dung dịch ZnSO4 16,1 % thu đc dung dịch A. Tính C% dung dịch A 07/08/2021 Bởi Amara Trộn lẫn 200g dung dịch KOH 7% với 100 dung dịch ZnSO4 16,1 % thu đc dung dịch A. Tính C% dung dịch A
$n_{KOH}=\dfrac{200.7\%}{56}=0,25(mol)$ $n_{ZnSO_4}=\dfrac{100.16,1\%}{161}=0,1(mol)$ $ZnSO_4+2KOH\to Zn(OH)_2+K_2SO_4$ $\Rightarrow KOH$ dư $n_{Zn(OH)_2}=0,1(mol)$ $n_{K_2SO_4}=0,1.2=0,2(mol)$ $n_{KOH\text{dư}}=0,25-0,1.2=0,05(mol)$ $Zn(OH)_2+2KOH\to K_2ZnO_2+2H_2O$ $\Rightarrow Zn(OH)_2$ dư $n_{K_2ZnO_2}=n_{Zn(OH)_2\text{pứ}}=0,025(mol)$ $\Rightarrow n_{Zn(OH)_2\text{dư}}=0,1-0,025=0,075(mol)$ $\Rightarrow m_{dd\text{spu}}=200+100-0,075.99=292,575g$ $C\%_{K_2SO_4}=\dfrac{0,2.174.100}{292,575}=11,89\%$ $C\%_{K_2ZnO_2}=\dfrac{0,025.175.100}{292,575}=1,5\%$ Bình luận
Đáp án: Giải thích các bước giải: nKOH=200.7%56=0,25(mol)nKOH=200.7%56=0,25(mol) nZnSO4=100.16,1%161=0,1(mol)nZnSO4=100.16,1%161=0,1(mol) ZnSO4+2KOH→Zn(OH)2+K2SO4ZnSO4+2KOH→Zn(OH)2+K2SO4 ⇒KOH⇒KOH dư nZn(OH)2=0,1(mol)nZn(OH)2=0,1(mol) nK2SO4=0,1.2=0,2(mol)nK2SO4=0,1.2=0,2(mol) nKOHdư=0,25−0,1.2=0,05(mol)nKOHdư=0,25−0,1.2=0,05(mol) Zn(OH)2+2KOH→K2ZnO2+2H2OZn(OH)2+2KOH→K2ZnO2+2H2O ⇒Zn(OH)2⇒Zn(OH)2 dư nK2ZnO2=nZn(OH)2pứ=0,025(mol)nK2ZnO2=nZn(OH)2pứ=0,025(mol) ⇒nZn(OH)2dư=0,1−0,025=0,075(mol)⇒nZn(OH)2dư=0,1−0,025=0,075(mol) ⇒mddspu=200+100−0,075.99=292,575g⇒mddspu=200+100−0,075.99=292,575g C%K2SO4=0,2.174.100292,575=11,89%C%K2SO4=0,2.174.100292,575=11,89% C%K2ZnO2=0,025.175.100292,575=1,5% Bình luận
$n_{KOH}=\dfrac{200.7\%}{56}=0,25(mol)$
$n_{ZnSO_4}=\dfrac{100.16,1\%}{161}=0,1(mol)$
$ZnSO_4+2KOH\to Zn(OH)_2+K_2SO_4$
$\Rightarrow KOH$ dư
$n_{Zn(OH)_2}=0,1(mol)$
$n_{K_2SO_4}=0,1.2=0,2(mol)$
$n_{KOH\text{dư}}=0,25-0,1.2=0,05(mol)$
$Zn(OH)_2+2KOH\to K_2ZnO_2+2H_2O$
$\Rightarrow Zn(OH)_2$ dư
$n_{K_2ZnO_2}=n_{Zn(OH)_2\text{pứ}}=0,025(mol)$
$\Rightarrow n_{Zn(OH)_2\text{dư}}=0,1-0,025=0,075(mol)$
$\Rightarrow m_{dd\text{spu}}=200+100-0,075.99=292,575g$
$C\%_{K_2SO_4}=\dfrac{0,2.174.100}{292,575}=11,89\%$
$C\%_{K_2ZnO_2}=\dfrac{0,025.175.100}{292,575}=1,5\%$
Đáp án:
Giải thích các bước giải:
nKOH=200.7%56=0,25(mol)nKOH=200.7%56=0,25(mol)
nZnSO4=100.16,1%161=0,1(mol)nZnSO4=100.16,1%161=0,1(mol)
ZnSO4+2KOH→Zn(OH)2+K2SO4ZnSO4+2KOH→Zn(OH)2+K2SO4
⇒KOH⇒KOH dư
nZn(OH)2=0,1(mol)nZn(OH)2=0,1(mol)
nK2SO4=0,1.2=0,2(mol)nK2SO4=0,1.2=0,2(mol)
nKOHdư=0,25−0,1.2=0,05(mol)nKOHdư=0,25−0,1.2=0,05(mol)
Zn(OH)2+2KOH→K2ZnO2+2H2OZn(OH)2+2KOH→K2ZnO2+2H2O
⇒Zn(OH)2⇒Zn(OH)2 dư
nK2ZnO2=nZn(OH)2pứ=0,025(mol)nK2ZnO2=nZn(OH)2pứ=0,025(mol)
⇒nZn(OH)2dư=0,1−0,025=0,075(mol)⇒nZn(OH)2dư=0,1−0,025=0,075(mol)
⇒mddspu=200+100−0,075.99=292,575g⇒mddspu=200+100−0,075.99=292,575g
C%K2SO4=0,2.174.100292,575=11,89%C%K2SO4=0,2.174.100292,575=11,89%
C%K2ZnO2=0,025.175.100292,575=1,5%