Bài 1:
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,1}{1}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,05\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
\(\Rightarrow V_{O_2\left(dư\right)}=0,05.22,4=1,12\left(l\right)\)
\(n_{H_2O}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{H_2O}=0,1.18=1,8\left(g\right)\)
Bài 2:
\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
\(n_{HCl}=\dfrac{1,46}{36,5}=0,04\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,05}{1}>\dfrac{0,04}{2}\), ta được Zn dư.
Theo PT: \(n_{Zn\left(pư\right)}=\dfrac{1}{2}n_{HCl}=0,02\left(mol\right)\Rightarrow n_{Zn\left(dư\right)}=0,05-0,02=0,03\left(mol\right)\)
\(\Rightarrow m_{Zn\left(dư\right)}=0,03.65=1,95\left(g\right)\)
\(PT:2H_2+O_2\rightarrow2H_2O\)
\(\dfrac{n_{O_2\left(ĐB\right)}}{n_{O_2\left(PT\right)}}=\dfrac{0,1}{1}>\dfrac{n_{H_2\left(ĐB\right)}}{n_{H_29PT}}=\dfrac{0,1}{2}\)
\(\rightarrow O_2\) dư H2 hết. Tính theo H2
Theo PT: \(n_{O_2\left(pứ\right)}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}0,1=0,05\left(mol\right)\)
\(\rightarrow n_{O_2\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
\(\rightarrow V_{O_2\left(dư\right)}=\dfrac{n}{22,4}=\dfrac{0,05}{22,4}\approx0,002\left(l\right)\)
Theo PT : \(n_{H_2O}=n_{H_2}=0,1\left(mol\right)\\ \rightarrow m_{H_2O}=n\cdot M=0,1\cdot18=1,8\left(g\right)\)
2, \(n_{Zn}=\dfrac{m}{M}=\dfrac{3,25}{65}=0,05\left(mol\right)\\ n_{HCl}=\dfrac{m}{M}=\dfrac{1,46}{36,5}=0,04\left(mol\right)\)
\(PT+Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(\dfrac{n_{Zn\left(ĐB\right)}}{n_{Zn\left(PT\right)}}=\dfrac{0,05}{1}>\dfrac{n_{HCl\left(ĐB\right)}}{n_{HCl\left(PT\right)}}=\dfrac{0,04}{2}\)
\(\rightarrow\) Zn dư , HCl hết . Tính theo HCl
Theo PT: \(n_{Zn\left(pứ\right)}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\cdot0,04=0,02\left(mol\right)\)
\(\rightarrow m_{Zn\left(pứ\right)}=n\cdot M=0,02\cdot65=1,3\left(g\right)\)
\(\rightarrow m_{Zn\left(dư\right)}=3,25-1,3=1,95\left(g\right)\)
