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a) \(NaOH+HCl\rightarrow NaCl+H_2O\)
b) \(m_{NaOH}=40.20\%=8g\Rightarrow n_{NaOH}=\dfrac{8}{40}=0,2mol\)
\(2NaOH\left(0,2\right)+H_2SO_4\left(0,1\right)\rightarrow Na_2SO_4+2H_2O\)
\(m_{H_2SO_4}=0,1.98=9,8g\Rightarrow m_{ddH_2SO_4}=\dfrac{100.9,8}{10}=98g\)
1.NaOH+HCl--->NaCl+H2O
nNaOH=(200.10%)/40=0,5
=>nHCl=nNaOH=0,5
=>mddHCl=(0,5.36,5)/3,65%=500 g
2:a,2NaOH+H2SO4−−>Na2SO4+H2O2
Theo pthh, ta có: nNaOH=2.nH2SO4=0,4mol
-->mNaOH=16g
-->md/dNaOH=80g
b, Ta có: nKOH=0,4mol
-->md/dKOH=400g
-->V=383ml
Thái Thùy Linh
nNaOH= 2*0.1=0.2 mol
PTHH: NaOH + HCl = NaCl + H2O
0.2............0.2
a) VHCl = 0.2/2= 0.1 mol
b) PTHH: 2NaOH + H2SO4 = Na2SO4 + 2H2O
0.2.............0.1
mH2SO4= 0.1*98=9.8g
=>mddH2SO4 = \(\dfrac{m_{ct}\cdot100}{C\%}\) =\(\dfrac{9.8\cdot100}{24.5}=\) 40g
a)$n_{H_2SO_4} = 0,3.1,5 = 0,45(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4} = 0,9(mol)$
$m_{dd\ NaOH} = \dfrac{0,9.40}{40\%} = 90(gam)$
b)
$n_{KOH} = n_{NaOH} = 0,9(mol0$
$m_{dd\ KOH} = \dfrac{0,9.56}{5,6\%} = 900(gam)$
$V_{dd\ KOH} = \dfrac{m}{D} = \dfrac{900}{1,045} = 861,24(ml)$
a) NaOH+HCl---->NaCl+H2O
n HCl=0,2.2=0,4(mol)
Theo pthh
n NaOH =n HCl =0,4(mol)
V NaOH= 0,4/0,1=4(l)=400ml
b) Ca(OH)2+2HCl---->CaCl2+2H2O
Theo pthhj
n Ca(OH)2=1/2 n HCl =0,2(mol)
m Ca(OH)2=\(\frac{0,2.74.100}{5}=296\left(g\right)\)
Bài 2
Ca(OH)2+2HCl---->CaCl2+2H2O
n HCl=0,2.2=0,4(mol)
Theo pthh
n Ca(OH)2=1/2 n HCl =0,2(mol)
m Ca(OH)2=\(\frac{0,2.74.200}{10}=148\left(g\right)\)
Bài 3
H2SO4+2NaOH--->Na2SO4+H2O
n H2SO4=0,2.1=0,2(mol)
Theo pthh
n NaOH =2n H2SO4=0,4(mol)
m NaOH=\(\frac{0,4.40.100}{20}=80\left(g\right)\)
Bài 4
HCl+NaOH---->NaCl+H2O
n HCl=0,2.1=0,2(mol)
Theo pthh
n NaCl =n HCl =0,2(mol)
m NaCl=0,2.58,5=11,7(g)
n NaOH =n HCl=0,2(mol)
m NaOH=\(\frac{0,2.40.100}{20}=40\left(g\right)\)
Câu 1:
\(\text{n hcl = 0,2.0,2 = 0,04 mol}\)
\(\text{a, naoh + hcl ---> nacl + h2o}\)
n naoh = n hcl = 0,04 mol
\(\Rightarrow\text{V naoh = 0,04 ÷ 0,1 = 0,4 lít --> V = 400ml}\)
b, \(\text{ca(oh)2 + 2hcl ---> cacl2 +2 h2o}\)
n ca(oh)2 =1/2. n hcl = 0 ,02 mol
\(\Rightarrow\text{--> m dd ca(oh)2 = 0,02. 74÷ 5 .100 = 29,6g}\)
Câu 2 :
\(\text{ n hcl = 0,2.2 = 0,4 mol}\)
\(\text{ca(oh)2 + 2hcl ---> cacl2 +2 h2o}\)
n ca(oh)2 =1/2. n hcl = 0 ,2 mol
\(\Rightarrow\text{m dd Ca(OH)2 = 0,2.74÷10.100 = 148g}\)
Câu 3:
\(\text{2NaOH + H2SO4 -> Na2SO4 + H2O}\)
Ta có : nH2SO4=0,2.1=0,2 mol
Theo ptpu: nNaOH=2nH2SO4=0,2.2=0,4 mol
\(\text{-> mNaOH=0,4.40=16 gam }\)
m dung dịch NaOH=16/20%=80 gam
Câu 4
\(\text{NaOH + HCl -> NaCl + H2O}\)
Ta có: nHCl=0,2.1=0,2 mol
Theo ptpu: nNaOH=nNaCl=nHCl=0,2 mol
\(\Rightarrow\text{mNaOH=0,2.40=8 gam}\)
\(\Rightarrow\text{m dung dịch NaOH=8/20%=40 gam}\)
muối là NaCl 0,2 mol -> mNaCl=0,2.58,5=11,7 gam
Duong Le buithianhtho Linh Bùi Lan Anh Hoàng Ngọc Anh Shizadon
Trần Hữu Tuyển Phùng Hà Châu Quang Nhân
Bài 2
\(n_{H_2SO_4}=0,5.0,12=0,06\left(mol\right)\)
PTHH: H2SO4 + 2NaOH --> Na2SO4 + 2H2O
______ 0,06 -----> 0,12____________________ (mol)
=> \(m_{NaOH}=0,12.40=4,8\left(g\right)\)
=> m dd NaOH = \(\frac{4,8.100}{12,5}=38,4\left(g\right)\)
b)
PTHH: H2SO4 + 2KOH --> K2SO4 + 2H2O
______ 0,06 -----> 0,12 _________________ (mol)
=> \(m_{KOH}=0,12.56=6,72\left(g\right)\)
=> m dd KOH = \(\frac{6,72.100}{11,2}=60\left(g\right)\)
=> V dd KOH = \(\frac{60}{1,12}=53,57\left(ml\right)\)
mHCl=7,3%.200=14,6(g) -> nHCl=0,4(mol)
a) PTHH: HCl + NaOH -> NaCl + H2O
Ta có: nNaOH=nNaCl=nHCl=0,4(mol)
=> mNaOH=0,4.40=16(g)
=>mddNaOH=16:10%=160(g)
=>a=160(g)
b) PTHH: Ba(OH)2 +2 HCl -> BaCl2 + 2 H2O
nBa(OH)2= nHCl/2=0,4/2=0,2(mol)
=> mBa(OH)2=0,2.171=34,2(g)
=>mddBa(OH)2= 34,2 : 17,1%=200(g)
=> Nếu thay dd NaOH bằng dung dịch Ba(OH)2 17,1% thì cần 200 gam dung dịch Ba(OH)2 em nhé!