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`1) C_[M_[HCl]] = [ 0,75 ] / [ 0,5 ] = 1,5 (M)`
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`2)n_[Ca(OH)_2] = 37 / 74 = 0,5 (mol)`
`-> C_[M_[Ca(OH)_2]] = [ 0,5 ] / [ 1,5 ] ~~ 0,33 (M)`
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`3) n_[NaOH] = 0,25 + 20 / 40 = 0,75 (mol)`
`-> C_[M_[NaOH]] = [ 0,75 ] / 2 = 0,375 (M)`
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`4) n_[H_2 SO_4] = 49 / 98 = 0,5 (mol)`
`-> C_[M_[H_2 SO_4]] = [ 0,5 ] / 2 = 0,25 (M)`
\(a,n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\ C_{M\left(HCl\right)}=\dfrac{0,2}{0,4}=0,5M\\ b,n_{H_2SO_4}=\dfrac{73,5}{98}=0,75\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,75}{0,5}=1,5M\\ n_{NaOH}=\dfrac{16}{40}=0,4\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,4}{0,25}=1,6M\\ n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\\ C_{M\left(Ba\left(OH\right)_2\right)}=\dfrac{0,2}{0,8}=0,25M\)
a) Ta có: \(n_{NaCl}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,1}{0,5}=0,2\left(M\right)=\left[Na^+\right]=\left[Cl^-\right]\)
b) Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\)
\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=0,4\left(M\right)\\\left[OH^-\right]=0,8\left(M\right)\end{matrix}\right.\)
c) Ta có: \(n_{H_2SO_4}=0,025\cdot2=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,125+0,025}\approx0,33\left(M\right)\) \(\Rightarrow\left\{{}\begin{matrix}\left[H^+\right]=0,66\left(M\right)\\\left[SO_4^{2-}\right]=0,33\left(M\right)\end{matrix}\right.\)
Bài 8. Trong 300 ml dd Ca(OH)2 có hoà tan 5,18 gam Ca(OH)2. Tính nồng độ mol của dung dịch thu được?
nCa(OH)2 = 0,07(mol)
=> CM Ca(OH)2 = n/V = 0,07 / 0,3 = 0,233(M)
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\\ C_{M_{NaOH}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)
n NaOH= \(\dfrac{4}{40}\)=0,1(mol)
C MnaoH=\(\dfrac{0,1}{0,4}\)=0,25(M)
a) $CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
Theo PTHH :
$n_{CO_2} = n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$
$V_{CO_2} = 0,1.22,4 = 2,24(lít)$
b) $n_{HCl} = 2n_{CaCO_3} = 0,2(mol)$
$C_{M_{HCl}} = \dfrac{0,2}{0,25} = 0,8M$
c) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,1(mol)$
$m_{CaCO_3} = 0,1.100 = 10(gam)$
a)\(n_{K_2O}=\dfrac{23,5}{94}=0,25mol\)
\(K_2O+H_2O\rightarrow2KOH\)
0,25 0,25 0,5
\(C_M=\dfrac{0,5}{0,5}=1M\)
b)Để trung hòa: \(n_{H^+}=n_{OH^-}=0,5\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{H^+}=0,25mol\)
\(m_{H_2SO_4}=0,25\cdot98=24,5g\)
\(\Rightarrow m_{ddHCl}=\dfrac{24,5\cdot100\%}{60\%}=\dfrac{245}{6}g\)
Thể tích dung dịch:
\(V=\dfrac{m}{D}=\dfrac{\dfrac{245}{6}}{1,5}\approx27,22ml\)
\(n_{K_2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\\ K_2O+H_2O\rightarrow2KOH\\ n_{KOH}=2.0,25=0,5\left(mol\right)\\ a,C_{M\text{dd}A}=C_{M\text{dd}KOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b,2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ n_{H_2SO_4}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{H_2SO_4}=0,25.98=24,5\left(g\right)\\ m_{\text{dd}H_2SO_4}=\dfrac{24,5.100}{60}=\dfrac{245}{6}\left(g\right)\\ V_{\text{dd}H_2SO_4}=\dfrac{\dfrac{245}{6}}{1,5}=\dfrac{245}{9}\left(ml\right)\approx27,222\left(ml\right)\)
\(n_{NaOH}=\dfrac{m_{NaOH}}{M_{NaOH}}=\dfrac{8}{36}=\dfrac{2}{9}\left(mol\right)\)
\(C_M=\dfrac{n}{V}=\dfrac{\dfrac{2}{9}}{0,05}=4\left(M\right)\)
\(1,C_{M\left(HCl\right)}=\dfrac{0,75}{0,5}=1,5M\\ 2,n_{Ca\left(OH\right)_2}=\dfrac{37}{74}=0,5\left(mol\right)\\ C_{M\left(Ca\left(OH\right)_2\right)}=\dfrac{0,5}{1,5}=0,33M\\ 3,n_{NaOH}=0,25+\dfrac{20}{40}=0,75\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,75}{2}=0,375M\\ 4,n_{H_2SO_4}=\dfrac{49}{98}=0,5\left(mol\right)\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,5}{2}=0,25M\)