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PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{1,2395}{24,79}=0,05\left(mol\right)\)
a, Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=0,05\left(mol\right)\Rightarrow m_{Na_2CO_3}=0,05.106=5,3\left(g\right)\)
b, \(n_{NaOH}=2n_{CO_2}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
n NaOH = 2 n CO 2 = 1,12x2 /22,4 = 0,1 (mol)
Nồng độ mol của dung dịch NaOH là 1M.
\(A,\) Tên khí A là SO2
\(B,\) \(n_{SO_2}=\frac{1.12}{22.4}=0.05\left(mol\right)\)
\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)
0.05 0.05 0.05
\(m_{H_2SO_4}=0.05\times98=4.9\left(g\right)\)
\(C\%_{H_2SO_4}=\frac{4.9}{24.5}\times100=20\%\)
\(C,\)
\(m_{K_2SO_3}=0.05\times158=7.9\left(g\right)\)
\(\%m_{K_2SO_3}=\frac{7.9}{21}\times100=37.6\%\)
\(\%m_{K_2SO_4}=100\%-37.6\%=62.4\%\)
a) \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
b) \(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{CO_2}=0,1\left(mol\right)\)
=> \(CM_{NaOH}=\dfrac{0,1}{0,1}=1M\)
c) Sửa đề DNaOH = 1,2g/ml
\(m_{ddsaupu}=0,05.44+100.1,2=122,2\left(g\right)\)
\(n_{Na_2CO_3}=n_{CO_2}=0,05\left(mol\right)\)
=> \(C\%_{Na_2CO_3}=\dfrac{0,05.106}{122,2}.100=4,34\%\)
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Ta có: \(n_{Na_2SO_4}=n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot\dfrac{10}{40}=0,125\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,125\cdot98}{10\%}=122,5\left(g\right)\\m_{Na_2SO_4}=0,125\cdot142=17,75\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{17,75}{10+122,5}\cdot100\%\approx13,4\%\)
nCO2= 2,24 :22,4= 0,1 Mol
a) lập pthh của pư
CO2 +2 NaOH -------->Na2CO3 + H2O
1mol 2mol 1mol 1mol
0,1mol 0,2mol 0,1mol 0,1mol
mCO2 =0,1 .44= 4,4 gam
mNa2CO3 = 0,1 . 106 =10,6 gam
mH2O= 0,1 . 18 = 1,8 gam
mdd sau pư = mct + m dung môi = 4,4+1,8 =6,2 gam
C%Na2CO3 =( mct /mdd ). 100% = (10,6 /6,2 ) .100% =170,97 %
b) CmNaOH = n/ v = 0,2 /0,1 =2 mol/lít
a) PTHH: ZnO + H2SO4 → ZnSO4 + H2O (1)
\(n_{ZnO}=\dfrac{6,48}{81}=0,08\left(mol\right)\)
b) Theo PT1: \(n_{H_2SO_4}=n_{ZnO}=0,08\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,08}{0,05}=1,6\left(M\right)\)
c) CO2 + 2NaOH → Na2CO3 + H2O (2)
\(n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(m_{NaOH}=35\times10\%=3,5\left(g\right)\)
\(\Rightarrow n_{NaOH}=\dfrac{3,5}{40}=0,0875\left(mol\right)\)
Theo PT2: \(n_{CO_2}=\dfrac{1}{2}n_{NaOH}\)
Theo bài: \(n_{CO_2}=\dfrac{4}{7}n_{NaOH}\)
Vì: \(\dfrac{4}{7}>\dfrac{1}{2}\) ⇒ CO2 dư ⇒ phản ứng tiếp
Na2CO3 + CO2 + H2O → 2NaHCO3 (3)
Theo PT2: \(n_{CO_2}pư=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,0875=0,04375\left(mol\right)\)
\(\Rightarrow n_{CO_2}dư=n_{CO_2\left(3\right)}=0,05-0,04375=0,00625\left(mol\right)\)
Theo PT2: \(n_{Na_2CO_3}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,0875=0,04375\left(mol\right)=n_{Na_2CO_3\left(3\right)}\)
\(\Rightarrow m_{Na_2CO_3}=0,04375\times106=4,6375\left(g\right)\)
Theo PT3: \(n_{Na_2CO_3}=n_{CO_2}\)
Theo bài: \(n_{Na_2CO_3}=7n_{CO_2}\)
Vì: \(7>1\) ⇒ Na2CO3 dư
Theo PT3: \(n_{NaHCO_3}=2n_{CO_2}=2\times0,00625=0,0125\left(mol\right)\)
\(\Rightarrow m_{NaHCO_3}=0,0125\times84=1,05\left(g\right)\)
Theo PT3: \(n_{Na_2CO_3}pư=n_{CO_2}=0,00625\left(mol\right)\)
\(\Rightarrow n_{Na_2CO_3}dư=0,04375-0,00625=0,0375\left(mol\right)\)
\(\Rightarrow m_{Na_2CO_3}dư=0,0375\times106=3,975\left(g\right)\)
\(\Rightarrow m_{Na_2CO_3}tt=4,6375-3,975=0,6625\left(g\right)\)