Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
nH2SO4 = 0.2*1=0.2 mol
2NaOH + H2SO4 --> Na2SO4 + H2O
0.4________0.2
mNaOH = 0.4*40=16g
2KOH + H2SO4 --> K2SO4 + H2O
0.4______0.2
mKOH= 0.4*56=22.4g
mddKOH = 22.4*100/5.6=400g
VddKOH = 400/1.045=382.77ml
\(n_{H_2SO_4}=0,2\times1=0,2\left(mol\right)\)
H2SO4 + 2NaOH → Na2SO4 + 2H2O (1)
a) Theo PT1: \(n_{NaOH}=2n_{H_2SO_4}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4\times40=16\left(g\right)\)
b) H2SO4 + 2KOH → K2SO4 + 2H2O (2)
Theo PT2: \(n_{KOH}=2n_{H_2SO_4}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{KOH}=0,4\times56=22,4\left(g\right)\)
\(\Rightarrow m_{ddKOH}=\frac{22,4}{5,6\%}=400\left(g\right)\)
\(\Rightarrow V_{ddKOH}=\frac{400}{1,045}=382,78\left(ml\right)\)
nH2SO4=0,02.1=0,02(mol)nH2SO4=0,02.1=0,02(mol)
PTHH: H2SO4+2NaOH→Na2SO4+2H2OH2SO4+2NaOH→Na2SO4+2H2O
pư..............0,02..........0,04..............0,02...........0,04 (mol)
⇒mNaOH=0,04.40=1,6(g)⇒mNaOH=0,04.40=1,6(g)
⇒mddNaOH(20%)=1,620%=8(g)⇒mddNaOH(20%)=1,620%=8(g)
PTHH: H2SO4+2KOH→K2SO4+2H2OH2SO4+2KOH→K2SO4+2H2O
pư............0,02............0,04............0,02..........0,04 (mol)
⇒mKOH=0,04.56=2,24(g)⇒mKOH=0,04.56=2,24(g)
⇒mddKOH(5,6%)=2,245,6%=40(g)⇒mddKOH(5,6%)=2,245,6%=40(g)
⇒VKOH=401,045≈38,28(ml)
\(n_{H_2SO_4}=0,1.0,75=0,075mol\\ H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)
0,075 0,15 0,075 0,15
\(a)m_{K_2SO_4}=0,075.175=13,05mol\)
\(b)H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ n_{NaOH}=0,075.2=0,15mol\\ m_{ddNaOH}=\dfrac{0,15.40}{15\%}\cdot100\%=40g\\ V_{ddNaOH}=\dfrac{40}{1,05}=38,1ml\)
a.250ml=0,25l ; nHCl=0,25.1,5=0,375mol
KOH+HCl->KCl+H2O
1mol 1mol 1mol
0,375 0,375 0,375
VKOh=0,375/2=0,1875l
b.CM KCL=0,375/0,25=1,5M
c.NaOH+HCL=NaCl+H2O
1mol 1mol
0,375 0,375
mdd NaOH=0,375.40.100/10=150g
a) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: CO2 + 2NaOH → Na2CO3 + H2O
Mol: 0,15 0,3 0,15
\(C_{M_{ddNaOH}}=\dfrac{0,3}{0,2}=1,5M\)
b) Na2CO3: natri cacbonat
\(m_{Na_2CO_3}=0,15.106=15,9\left(g\right)\)
c)
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,15 0,075
\(V_{ddH_2SO_4}=\dfrac{0,075}{1}=0,075\left(l\right)=75\left(ml\right)\)
a) nH2SO4 = 0,3.1,5 = 0,45 mol
H2SO4 + 2NaOH --> Na2SO4 + 2H2O
0,45 ---> 0,9
=>mNaOH = 0,9.40 = 36 g
=>mdd NaOH = (36.100)/40 = 90 g
b) H2SO4 + 2KOH --> K2SO4 + 2H2O
0,45 -----> 0,9
=>mKOH = 0,9.56 = 50,4 g
=> mddKOH = (50,4.100)/5,6 = 900 g
A) nH2SO4=0,3.1,5=0,45 (mol)
H2SO4+ 2NaOH➝ Na2SO4+H2O
0,45 0,9 0,45 (mol)
mNaOH = 0,9.40 = 36 (g)
m dd NaOH= \(\frac{36.100}{40}\) = 90 (g)
B) Không hiểu đề 
..
nH2SO4=0,02.1=0,02(ol)
a) PTHH: 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O
0,04____________0,02____0,02(mol)
mNaOH=0,04.40= 1,6(g)
=>mddNaOH= (1,6.100)/20= 8(g)
b) PTHH: H2SO4 + 2 KOH -> K2SO4 + 2 H2O
0,2____________0,04(mol)
=>mKOH=0,04.56=2,24(g)
=>mddKOH= (2,24.100)/5,6=40(g)
=>VddKOH= mddKOH/DddKOH= 40/1,045=38,278(ml)
a, nH2SO4=0.02*1=0.02(mol)
H2SO4 + NaOH ➞ Na2SO4 +H2O
0.02.........0.02........0.02.........0.02.......(mol)
m dung dịch NaOH=(0.02*40)*100/20=4(g)
b) H2SO4 + KOH ➞ K2SO4 +H2O
....0.02.......0.02..........0.02......0.02...(mol)
mdung dịch KOH=(0.02*56)*100/5.6=20(g)
Vdung dịch=20/1.045=19.139(ml)
nH2SO4=0,1(mol)
a) PTHH: 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O
b) 0,2___________0,1________0,1(mol)
mNaOH=0,2.40=8(g)
=>mddNaOH=(8.100)/25= 32(g)
c) mNa2SO4=0,1.142=14,2(g)
d) PTHH: 2 KOH + H2SO4 -> K2SO4 +2 H2O
nKOH=0,2(mol) => mKOH=0,2.56=11,2(g)
=> mddKOH=(11,2.100)/8=140(g)
=> VddKOH= 140/1,085=129,03(ml)
Chúc em học tốt!
thank you