a) PTHH: Zn + 2HCl -> ZnCl2 + H2 (1)
Ta có: \(m_{HCl}=\dfrac{7,3.50}{100}=3,65\left(g\right)\\ =>n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ =>n_{ZnCl_2}=n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ =>m_{ZnCl_2}=0,05.136=6,8\left(g\right)\)
b) PTHH: H2 + CuO -to-> Cu + H2O (2)
Ta có: \(n_{H_2\left(2\right)}=n_{H_2\left(1\right)}=0,05\left(mol\right)\\ =>n_{Cu\left(2\right)}=n_{H_2\left(2\right)}=0,05\left(mol\right)\\ =>m_{Cu}=0,05.64=3,2\left(g\right)\)
a: PTHH: Zn + 2HCl --> ZnCl2 + H2(1)
mHCl = 50.7,3%= 3,65g
nHCL = 3,65/36,5=0,1 mol
Theo PTHH nZnCl2 = 1/2nHCl = 0,05 mol
=> mZnCl2 = 0,5.136=6,8g
Theo PTHH nH2 =1/2nHCl= 0,05 mol
b: PTHH: H2 + CuO --> Cu+ H2O (2)
nH2(1) = nH2 (2) = 0,05 mol
Theo PTHH(2) nCu= nH2= 0,05 mol
=>mCu = 0,05.64=3,2g
chúc bạn học tốt :))
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
a)
Theo đề bài ta có :
mHCl = 50 . 7,3% = 3,65(g)
=> nHCl = 3,65 : 36,5 = 0,1 (mol)
=> nZnCl2 = 0,1 : 2 = 0,05(mol)
=> mZnCl2 = 0,05 . 136 = 6,8(g)
b)
Ta có : nH2 = 0,05 (mol)
H2 + CuO \(\rightarrow\) Cu + H2O
Ta thấy :
nH2 = nCu = 0,05(mol)
=> mCu = 0,05 . 64 = 3,2 (g)
