Trộn 500ml dd NaOH 5M với 200ml dd NaOH 30% (d=1,33 g /ml). Tính Cm của dd thu được
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13 tháng 9 2021
\(b.n_{NaOH\left(tổng\right)}=0,4.0,5+\dfrac{100.1,33.20\%}{40}=0,865\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\left(sau\right)\right]=\dfrac{0,865}{0,4+0,1}=1,73\left(M\right)\\ c.n_{HCl}=0,05.0,12=0,006\left(mol\right)\\ n_{HNO_3}=0,15.0,1=0,015\left(mol\right)\\ \left[H^+\right]=\dfrac{0,006+0,015}{0,05+0,15}=0,105\left(M\right)\\ \left[NO^-_3\right]=\dfrac{0,015}{0,05+0,15}=0,075\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,006}{0,05+0,15}=0,03\left(M\right)\)
\(d.n_{H_2SO_4}=0,4.0,05=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H^+\right]=\dfrac{0,02.2+0,07}{0,05+0,35}=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ f.n_{KOH}=\dfrac{20.1,31.32\%}{56}=\dfrac{131}{875}\left(mol\right)\\ n_{Ba\left(OH\right)_2}=0,08.1=0,08\left(mol\right)\\ \left[OH^-\right]=\dfrac{\dfrac{131}{875}+0,08.2}{0,02+0,08}=\dfrac{542}{175}\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,08}{0,02+0,08}=0,8\left(M\right)\)
\(\left[K^+\right]=\dfrac{\dfrac{131}{875}}{0,02+0,08}=\dfrac{262}{175}\left(M\right)\)
16 tháng 5 2020
\(\left\{{}\begin{matrix}n_{NaOH\left(dd.1M\right)}=0,3\left(mol\right)\\n_{NaOH\left(dd.1,5M\right)}=0,2.1,5=0,3\left(mol\right)\end{matrix}\right.\)
\(n_{NaOH\left(dd.sau\right)}=n_{NaOH\left(dd.1M\right)}+n_{NaOH\left(dd.1,5M\right)}=0,3+0,3=0,6\left(mol\right)\)
\(V_{dd\left(sau\right)}=300+200=500\left(ml\right)=0,5\left(l\right)\)
\(\Rightarrow CM_{dd\left(sau\right)}=\frac{0,6}{0,6}=1,2M\)
\(\left\{{}\begin{matrix}m_{dd.sau}=500.1,05=525\left(g\right)\\m_{NaOH}=06.40=24\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{Dd\left(spu\right)}=\frac{24}{525}.100\%=4,57\%\)
9 tháng 3 2022
nHCl = 1 . 0,2 = 0,2 (mol)
nNaOH = 1,2 . 0,3 = 0,36 (mol)
PTHH: NaOH + 2HCl -> NaCl + H2O
LTL: 0,36 > 0,2 => NaOH dư
nNaCl = nHCl = 0,2 (mol)
CMNaCl = 0,2/0,5 = 0,4M
LD
1
16 tháng 9 2021
\(V_{ddNaOH\left(tổng\right)}=400+200=600\left(ml\right)=0,6\left(l\right)\\ n_{NaOH\left(tổng\right)}=0,4.0,5+0,2.1,5=0,5\left(mol\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,5}{0,6}\approx0,833\left(M\right)\)
NM
1
6 tháng 7 2019
\(n_{HCl.5M}=0,05\times5=0,25\left(mol\right)\)
\(m_{ddHCl.30\%}=200\times1,33=266\left(g\right)\)
\(\Rightarrow m_{HCl.30\%}=266\times30\%=79,8\left(g\right)\)
\(\Rightarrow n_{HCl.30\%}=\frac{79,8}{36,5}=\frac{798}{365}\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}mới=0,25+\frac{798}{365}=\frac{3557}{1460}\left(mol\right)\)
\(\Sigma V_{ddHCl}mới=50+200=250\left(ml\right)=0,25\left(l\right)\)
\(\Rightarrow C_{M_{HCl}}mới=\frac{3557}{1460}\div0,25=9,75\left(M\right)\)
18 tháng 9 2021
BT1:
\(m_{NaCl}=50.20\%+150.10\%=25\left(g\right)\)
\(m_{ddNaCl}=50+150=200\left(g\right)\)
\(C\%_{ddNaCl}=\dfrac{25.100\%}{200}=12,5\%\)
BT2:
\(n_{H_2SO_4}=0,2.5+0,2.3=1,6\left(mol\right)\)
\(V_{ddH_2SO_4}=0,2+0,2=0,4\left(l\right)\)
\(C_{M_{ddH_2SO_4}}=\dfrac{1,6}{0,4}=4M\)
C
23 tháng 5 2017
Tính C% của dung dịch thu được:
Ta có: md d NaOH(1)=V.D=500.1,2=600(g)
Vd d NaOH(1)=500ml=0,5 (lít)
=> nNaOH(1)=CM.V=2.0,5=1 (mol)
=> mNaOH(1)=nNaOH.M=1.40=40(gam)
Ta có: md d NaOH(2)=V.D=300.1,1=330(g)
Vd d NaOH(2)=300ml=0,3 (lít)
=> nNaOH(2)=CM.V=0,5.0,3=0,15(mol)
=> mNaOH(2)=n.M=0,15.40=6(gam)
=> mNaOH mới=mNaOH(1) + mNaOH(2)=40+6=46(gam)
md d NaOH mới=md d NaOH(1) + md d NaOH(2)=600+330=930(gam)
=> \(C\%_{ddsauphanung}=\dfrac{m_{NaOHmới}.100\%}{m_{ddNaOHmoi}}=\dfrac{46.100}{930}\approx4,95\left(\%\right)\)
Tính CM của dung dịch thu được :
Ta có: Vd d NaOH mới=Vd d NaOH(1) + Vd d NaOH(2)=0,5+0,3=0,8(lít)
nd d NaOH mới= n d d NaOH(1) + n d d NaOH(2)= 1 + 0,15=1,15(mol)
=> \(C_M=\dfrac{n}{V}=\dfrac{1,15}{0,8}\approx1,44\left(M\right)\)
GH
13 tháng 7 2023
\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,2------->0,1--------->0,1
Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư
Trong dung dịch D có:
\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)
b
\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)
0,35<---------0,35
\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)
mNaOH trong dd 2 =200*1,33*30/100=79,8g
suy ra nNaOH= 0,5*5+79,8/40=4,495mol
Vdd=0,5+0,2=0,7(l)
suy ra Cmdd=4,495/0,7=899/140M