CaO + H2SO4 → CaSO4 + H2O
\(n_{CaO}=\frac{2,8}{56}=0,05\left(mol\right)\)
\(m_{H_2SO_4}=140\times2\%=2,8\left(g\right)\)
\(\Rightarrow m_{H_2SO_4}=\frac{2,8}{98}=\frac{1}{35}\left(mol\right)\)
a) Theo pT: \(n_{CaO}=n_{H_2SO_4}\)
Theo bài: \(n_{CaO}=\frac{7}{4}n_{H_2SO_4}\)
Vì \(\frac{7}{4}>1\) ⇒ CaO dư
Theo pT: \(n_{CaO}pư=n_{H_2SO_4}=\frac{1}{35}\left(mol\right)\)
\(\Rightarrow n_{CaO}dư=0,05-\frac{1}{35}=\frac{3}{140}\left(mol\right)\)
\(\Rightarrow m_{CaO}dư=\frac{3}{140}\times56=1,2\left(g\right)\)
b) Theo pT: \(n_{CaSO_4}=n_{H_2SO_4}=\frac{1}{35}\left(mol\right)\)
\(\Rightarrow m_{CaSO_4}=\frac{1}{35}\times136=3,89\left(g\right)\)
c) \(m_{dd}saupư=2,8+140-1,2=141,6\left(g\right)\)
\(\Rightarrow C\%_{CaSO_4}=\frac{3,89}{141,6}\times100\%=2,75\%\)
\(CaO+H_2SO_4\) ➞\(CaSO_4+H_2O\)
1 1 1 1
0,05 1,43
0,05 0,05
\(nCaO=\frac{m}{M}=\frac{2,8}{56}=0,05\left(mol\right)\)
\(n_{H_2SO_4}=\frac{m}{M}=\frac{140}{98}=1,43\left(mol\right)\)
So sánh:
\(n_{CaO}=\frac{0,05}{1}=0,05\left(mol\right)\) < \(n_{H_2SO_4}=\frac{1,43}{1}=1,43\left(mol\right)\)
⇒ \(H_2SO_4\) dư
⇒\(n_{H_2SO_4}\) tham gia là 0,05 mol
➞ \(n_{H_2SO_4}\)dư : 1,43 -0,05 =1,38(mol)
\(m_{CaSO_4}=n.M=0,05.136=6,8\left(g\right)\)
\(m_{H_2O}=n.M=0,05.18=0,9\left(g\right)\)
