a) Zn + 2HCl _____> ZnCl2 + H2
mZn + mHCl = mZnCl2 + mH2
=> mHCl = mZnCl2 + mH2 - mZn = 136 + 2 - 65 = 73 (g)
Câu 3:
\(a,Zn+2HCl\to ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{65}{65}=1(mol)\\ \Rightarrow n_{HCl}=2(mol)\\ \Rightarrow m_{HCl}=2.36,5=73(g)\)
\(b,Al_2^{x}O_3^{II}\\ \Rightarrow 2x=3.II\Rightarrow x=3\\ \Rightarrow Al(III)\\ \%_{Fe(FeO)}=\dfrac{56}{56+16}.100\%=77,78\%\)
Câu 4:
\(a,Zn+2HCl\to ZnCl_2+H_2\\ b,Fe_2O_3+3H_2\to 2Fe+3H_2O\\ c,4K+O_2\xrightarrow{t^o}2K_2O\\ d,2Al+3FeSO_4\to Al_2(SO_4)_3+3Fe\)
Tính:
\(c,n_{Cu}=\dfrac{6,4}{64}=0,1(mol)\\ d,n_{N_2}=\dfrac{5,6}{28}=0,2(mol)\\ \Rightarrow V_{N_2}=0,2.22,4=4,48(l)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,BTKL:m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ \Rightarrow m_{HCl}=272+4-130=146(g)\)