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Câu 1:
Sửa đề: 250ml NaCl 2 mol/l
Ta có: \(n_{NaCl}=0,25\cdot2=0,5\left(mol\right)\) \(\Rightarrow C_{M_{NaCl\left(sau\right)}}=\dfrac{0,5}{0,15+0,25}=1,25\left(M\right)\)
a) 300 ml nước ~ 300g nước
m dd = 25+300 = 325(g)
n CaCl2.6H2O = n CaCl2 = 25/219 = 0.11 (mol)
m CaCl2 = 0.11*111 = 12.21(g)
C%dd = 12.21/325*100% = 3.76%
V dd = 325/1.08 = 300.93(ml) = 0.3(l)
CM = 0.11/0.3 = 0.37M
ta có: nNa= 4,6/ 23= 0,2( mol)
PTPU
2Na+ 2H2O----> 2NaOH+ H2
0,2.....0,2..................................
theo gt: nH2O= 4,5/ 18= 0,25(mol)> 0,2 mol
=> H2O dư
theo PTPU ta có nH2= 1/2 nNa= 0,1( mol)
ADĐLBTKL ta có
mNaOH= mNa+ mH2O- mH2
= 4,6+ 4,5- 0,1. 2
= 8,9( g)
theo PTPU: nNaOH= nNa= 0,2( mol)
=> C%NaOH= 0,2. 40/ 8,9 . 100%= 89,89%
ta có VNaOH= 8,9/ 1,05= 8,48( ml)= 0,00848( l)
=> CM NaOH= 0,2/ 00848= 23,59M
\(n_{CO_2}=\frac{8,8}{44}=0,2\left(mol\right)\)
\(m_{NaOH}=200\times10\%=20\left(g\right)\)
\(\Rightarrow n_{NaOH}=\frac{20}{40}=0,5\left(mol\right)\)
CO2 + 2NaOH → Na2CO3 + H2O (1)
Theo pt1: \(n_{CO_2}=\frac{1}{2}n_{NaOH}\)
Theo bài: \(n_{CO_2}=\frac{2}{5}n_{NaOH}\)
Vì \(\frac{2}{5}< \frac{1}{2}\) ⇒ NaOH dư
Ta có: \(m_{dd}saupư=8,8+200=208,8\left(g\right)\)
Theo pT: \(n_{NaOH}pư=2n_{CO_2}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow n_{NaOH}dư=0,5-0,4=0,1\left(mol\right)\)
\(\Rightarrow m_{NaOH}dư=0,1\times40=4\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{4}{208,8}\times100\%=1,92\%\)
Theo Pt: \(n_{Na_2CO_3}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Na_2CO_3}=0,2\times106=21,2\left(g\right)\)
\(\Rightarrow C\%_{Na_2CO_3}=\frac{21,2}{208,8}\times100\%=10,15\%\)
Ta có: \(m_{dd}=300\cdot1,05=315\left(g\right)\) \(\Rightarrow C\%_{Na_2CO_3}=\dfrac{15,9}{315}\cdot100\%\approx5,05\%\)
Mặt khác: \(n_{Na_2CO_3}=\dfrac{15,9}{106}=0,15\left(mol\right)\) \(\Rightarrow C_{M_{Na_2CO_3}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)
Ta có : \(n_C:n_S=2:1->\dfrac{1}{2}n_c=n_S\)
Lại có : \(m_C+m_S=5,6\)
-> \(n_C.12+n_S.32=5,6\)
=> \(n_C.12+\dfrac{1}{2}n_C.32=5,6\)
=> \(n_C=0,2\left(mol\right)\)
-> \(n_S=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
PTHH : \(C+O_2\underrightarrow{t^o}CO_2\) (1)
\(S+O_2\underrightarrow{t^o}SO_2\) (2)
Từ (1) -> \(n_C=n_{O_2}=0,2\left(mol\right)\)
-> \(V_{O_2\left(1\right)}=0,2.22,4=4,48\left(l\right)\)
Từ (2) -> \(n_S=n_{O_2}=0,1\left(mol\right)\)
\(V_{O_2\left(2\right)}=0,1.22,4=2,24\left(l\right)\)
=> \(V=\dfrac{V_{O_2\left(1\right)}+V_{O_2\left(2\right)}}{20\%}=33,6\left(l\right)\)