Question

Giúp mình với

Answer 1

\(n_{Zn}=\dfrac{16.25}{65}=0.25\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.25...............................0.25\)

\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(n_{CuO}=\dfrac{32}{80}=0.4\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(1...........1\)

\(0.4.........0.25\)

\(LTL:\dfrac{0.4}{1}>\dfrac{0.25}{1}\Rightarrow CuOdư\)

\(n_{Cu}=n_{H_2}=0.25\left(mol\right)\)

\(m_{Cr}=m_{CuO\left(dư\right)}+m_{Cu}=\left(0.4-0.25\right)\cdot80+0.25\cdot64=28\left(g\right)\)

Answer 2

a, PTHH: Zn + 2HCl -> ZnCl2 + H2

b, nZn = m/M = 0,25mol

Theo PTHH : nH2 = nZn = 0,25 mol

=> V = 5,6l

c, PTHH : H2 + CuO -> Cu + H2O

...............0,25.....0,25....0,25.....

=> Sau phản ứng thu được 0,25mol Cu và 0,15mol CuO

=> m = mCu + mCuO = 28G