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: \(n_{Na_2CO_3}pư=n_{CO_2}=0,00625\left(mol\right)\)
\(\Rightarrow n_{Na_2CO_3}dư=0,04375-0,00625=0,0375\left(mol\right)\)
\(\Rightarrow m_{Na_2CO_3}dư=0,0375\times106=3,975\left(g\right)\)
\(\Rightarrow m_{Na_2CO_3}tt=4,6375-3,975=0,6625\left(g\right)\)
