Mg + H2SO4 ==> MgSO4 + H2
x..................................x...........x
Zn + H2SO4 ==> ZnSO4 + H2
y..................................y.........y
24x+65y=8,9
x+y= 0,2
=> x=y=0,1
=> a=28,1g
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\Leftrightarrow m_{H_2}=n.M=0,2\times2=0,4\left(g\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
|___| + |_____| = |_____| + |___|
8,9g + 19,6g = a + 0,4g
Áp dụng ĐLBTKL ta có:
\(m_{Zn+Mg}+m_{\left(H_2SO_4\right)}=m_a+m_{\left(H_2\right)}\)
\(\Leftrightarrow m_a=m_{Zn+Mg}+m_{H_2SO_4}-m_{H_2}=8,9+19,6-0,4=28,1\left(g\right)\)
nH2=\(\dfrac{v}{22,4}\)=\(\dfrac{44,8}{22,4}\)=0,2(mol)⇔mH2=n.M=0,2×2=0,4(g)
Mg+H2SO4→MgSO4+H2
Zn+H2SO4→ZnSO4+H2
|___| + |_____| = |_____| + |___|
8,9g + 19,6g = a + 0,4g
Áp dụng ĐLBTKL ta có:
mZn+Mg+m(H2SO4)=ma+m(H2)
⇔ma=mZn+Mg+mH2SO4−mH2=8,9+19,6−0,4=28,1(g)
PTHU
+ Mg + H2SO4 -> MgSO4 + H2
+ Zn + H2SO4 -> ZnSO4 + H2