\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{13}{65}=0,2(mol)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ c) CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{CuO} = n_{H_2} = 0,2(mol)\\ m_{CuO} = 0,2.80 = 16(gam)\)
a,PTHH: Zn+H2SO4-----ZnSO4 +H2
nZn = 0,2 mol
=> VH2=4,48 l
CuO + H2 ----- Cu +H2O
nH2= 0,2 mol
=> mCuO= 16g
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PT: \(n_{CuO}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)
Bạn tham khảo nhé!
\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(0,2mol\) \(Zn\rightarrow0,2molH_2\)
V\(H_2=0,2.22,4=4,48l\)
a/ PTHH: \(Zn+H_2SO_4\longrightarow ZnSO_4+H_2\)
b/ \(n_{Zn}=\dfrac{13}{65}=0,2(mol)\)
\(\to n_{H_2}=0,2(mol)\)
\(\to V_{H_2}=0,2.22,4=4,48(l)\)
c/ \(CuO+H_2\longrightarrow H_2O+Cu\)
\(n_{H_2}=0,2(mol)\)
\(\to n_{CuO}=0,2(mol)\)
\(\to m_{CuO}=0,2.80=16(g)\)
