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a) \(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)
\(n_{NaOH}=\dfrac{8\%.500}{40}=1\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
0,5---------------->0,5------->0,25
\(\Sigma n_{NaOH}=0,5+1=1,5\left(mol\right)\)
\(m_{ddsaupu}=11,5+500-0,25.2=511\left(g\right)\)
=> \(C\%_{NaOH}=\dfrac{1,5.40}{511}.100=11,74\%\)
b) Gọi thể tích dung dịch X cần tìm là V
\(n_{H^+}=V.1+V.0,5.1=2V\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
Ta có : \(n_{H^+}=n_{OH^-}=1,5\left(mol\right)\)
=> 2V=1,5
=> V=0,75(lít)
a, \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,2.84}{64,8}.100\%\approx25,93\%\\\%m_{MgSO_4}\approx74,07\%\end{matrix}\right.\)
b, - Dung dịch C gồm: MgCl2, MgSO4 và HCl dư.
Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{CO_2}=0,2\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{CO_2}=0,4\left(mol\right)\end{matrix}\right.\)
Ta có: \(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)
\(m_{MgSO_4}=64,8-0,2.84=48\left(g\right)\Rightarrow n_{MgSO_4}=\dfrac{48}{120}=0,4\left(mol\right)\)
Có: m dd sau pư = 64,8 + 100 - 0,2.44 = 156 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{156}.100\%\approx12,18\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{156}.100\%\approx2,34\%\\C\%_{MgSO_4}=\dfrac{48}{156}.100\%\approx30,77\%\end{matrix}\right.\)
c, PT: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(HCl+NaOH\rightarrow NaCl+H_2O\)
\(MgSO_4+2NaOH\rightarrow Na_2SO_4+Mg\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}+n_{MgSO_4}=0,6\left(mol\right)\)
\(\Rightarrow m_{cr}=m_{MgO}=0,6.40=24\left(g\right)\)
B1:
2NaOH+H2SO4\(\rightarrow\)Na2SO4+2H2O
nNaOH=\(\frac{4}{40}=0.1\)mol
=>nH2SO4=\(\frac{1}{2}\)nNaOH=0.05 mol
=>CM=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10-4 (M)
B2:
Mg+\(\frac{1}{2}\)O2\(\underrightarrow{t^0}\)MgO (1)
MgO+2HCl\(\rightarrow\)MgCl2+H2O (2)
nMg(1)=\(\frac{0,36}{24}=0,015mol\)
=>nMgO(1)=0,015=nMgO(2)
nHCl(2)=2nMgO(2)=0,03mol
=>CM(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)
nNa2O=0.1(mol)
pthh:Na2O+H2O->2NaOH
theo pthh:nH2O=nNa2O->nH2O=0.1(mol)->mH2O=0.1*18=1.8(g)
b)Theo pthh:nNaOH=2 nNa2O
->nNaOH=0.1*2=0.2(mol)
CM=0.2:0.2=1(M)
Khk bt là gì bạn? =)) câu b mình không tìm thấy lỗi sai.bạn chỉ cho mình với?
1) \(m_{ddNaCl}=50+200=250\left(g\right)\)
\(C\%_{NaCl}=\frac{50}{250}\times100\%=20\%\)
2) \(n_{NaOH}=0,2\times2=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4\times40=16\left(g\right)\)
3) 2KMnO4 \(\underrightarrow{to}\) K2MnO4 + MnO2 + O2
\(n_{O_2}=\frac{2,241}{22,4}=0,1\left(mol\right)\)
Theo pT: \(n_{KMnO_4}=2n_{O_2}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{KMnO_4}=0,2\times158=31,6\left(g\right)\)
4) \(n_{CaCl_2}=\frac{2,24}{111}=0,02\left(mol\right)\)
\(C_{M_{CaCl_2}}=\frac{0,02}{0,1}=0,2\left(M\right)\)
5) Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
\(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo pt1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=16-2,4=13,6\left(g\right)\)
\(\Rightarrow n_{MgO}=\frac{13,6}{40}=0,34\left(mol\right)\)
Theo pT1: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)
Theo pt2: \(n_{MgCl_2}=n_{MgO}=0,34\left(mol\right)\)
\(\Rightarrow\Sigma n_{MgCl_2}=0,1+0,34=0,44\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,44\times95=41,8\left(g\right)\)
6) Gọi CTHH là FexOy
Ta có: \(56x\div16y=70\div30\)
\(\Rightarrow x\div y=\frac{70}{56}\div\frac{30}{16}\)
\(\Rightarrow x\div y=2\div3\)
Vậy CTHH là Fe2O3
7) Na2O + H2O → 2NaOH
\(m_{ddNaOH}=155+145=300\left(g\right)\)
\(n_{Na_2O}=\frac{155}{62}=2,5\left(mol\right)\)
Theo pT: \(n_{NaOH}=2n_{Na_2O}=2\times2,5=5\left(mol\right)\)
\(\Rightarrow m_{NaOH}=5\times40=200\left(g\right)\)
\(C\%_{NaOH}=\frac{200}{300}\times100\%=66,67\%\)
Tham khảo
https://hoc247.net/cau-hoi-hoa-tan-naoh-ran-vao-nuoc-de-tao-thanh-2-dung-dich-a-va-b--qid95961.html
Câu 1
Na2O+ H2O ----->2NaOH
nNaOH=2.5 mol
Theo pt nNaOH=2nNa2O=5 mol
mdd=155+145=300 (g)
=>C%NaOH=\(\dfrac{5\cdot40\cdot100}{300}\)=66.67%
Câu 2
Gọi a(g) là khối lượng NaOH cần thêm vào
Theo đề bài ta có \(\dfrac{a}{a+200}\)=0.08
<=>0.08a+16=a
<=>0.92a=16=>a=17.39 (g)
Câu 4
C%=\(\dfrac{50\cdot100}{200+50}\)=20%