a) PTHH : \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
---------0,2------0,4--------0,2---------0,2 ( mol )
b) nZn= 13/65=0,2 mol
theo PTHH => nH2 = nZn =0,2 mol
=> VH2 = 0,2.22,4=4,48 lít
c) PTHH : \(CuO+H_2-nhiet->Cu+H_2O\)
------------1------------1-------------------1----------1 ( mol )
nCuO = 12/80=0,15 mol
ta có \(\dfrac{0,2}{1}>\dfrac{0,15}{1}=>H_2dư\)
nH2 dư = 0,2 - 0,15 =0,05 mol
=> khối lượng H2 dư = 0,05.2=0,1 gam
vậy...
a, PTHH: Zn+ 2HCl -> ZnCl2 + H2 (a)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
b, Ta có: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
c, PTHH: H2 + CuO -to-> Cu + H2O (b)
Từ đề bài, PTHH (a) và (b):
=> \(n_{H_2\left(b\right)}=n_{H_2\left(a\right)}=0,2\left(mol\right)\\ n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\\ =>\dfrac{n_{CuO\left(đề\right)}}{n_{CuO\left(PTHH\right)}}=\dfrac{0,15}{1}< \dfrac{n_{H_2\left(đề\right)}}{n_{H_2\left(PTHH\right)}}=\dfrac{0,2}{1}\)
=> CuO hết, H2 dư, tính theo CuO.
=> \(n_{H_2\left(f.ứng\right)}=n_{CuO}=0,15\left(mol\right)\\ =>n_{H_2\left(dư\right)}=0,2-0,15=0,05\left(mol\right)\\ =>m_{H_2\left(dư\right)}=0,05.2=0,1\left(g\right)\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a, PTHH: Zn + 2HCl -> ZnCl2 + H2 (1)
b, Ta có: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
=> \(V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
c, PTHH: H2 + CuO -to-> Cu + H2O (2)
\(n_{H_2\left(2\right)}=n_{H_2\left(1\right)}=0,2\left(mol\right)\\ n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
=> \(n_{H_2\left(b\right)}=n_{H_2\left(a\right)}=0,2\left(mol\right)\\ n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\\ =>\dfrac{n_{CuO\left(đề\right)}}{n_{CuO\left(PTHH\right)}}=\dfrac{0,15}{1}< \dfrac{n_{H_2\left(đề\right)}}{n_{H_2\left(PTHH\right)}}=\dfrac{0,2}{1}\)
=> CuO hết, H2 dư, tính theo CuO.
\(n_{H_2\left(f.ứng\right)}=n_{CuO}=0,15\left(mol\right)\\ =>n_{H_2\left(dư\right)}=0,2-0,15=0,05\left(mol\right)\\ =>m_{H_2\left(dư\right)}=0,05.2=0,1\left(g\right)\)