\(n_{SO_2}=\dfrac{2,24}{22,4}=0,1mol\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Cu}=y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}24x+64y=4,4\\BTe:2x+2y=2n_{SO_2}=2\cdot0,1=0,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
\(m_{Mg}=0,05\cdot24=1,2g\)
\(m_{Cu}=0,05\cdot64=3,2g\)
\(n_{H_2SO_4}=2n_{Mg}+2n_{Cu}=2\cdot0,05+2\cdot0,05=0,2mol\)
\(m_{H_2SO_4}=0,2\cdot98=19,6g\)
a)
Đặt nMg=x(mol),nCu=y(mol)
→24x+64y=4,4(g) (1)
Bảo toàn e : 2nMg+2nCu=2nSO2=0,2(mol)
→x+y=0,1(mol) (2)
Từ (1)(2) giải được:
x=y=0,05(mol)
mMg=0,05.24=1,2(g)
mCu=0,05.64=3,2(g)
b)
nH2SO4=2nSO2=0,2(mol)
mH2SO4=0,2.98=19,6(g)
Đặt nMg=x(mol);nCu=y(mol)
→24x+64y=4,4(1)
nSO2=\(\dfrac{2,24}{22,4}\)=0,1(mol)
Bảo toàn e: x+y=nSO2=0,1(2)
Từ (1)(2)→x=y=0,05(mol)
=>mMg=0,05.24=1,2(g)
mCu=0,05.64=3,2(g)
b)
Mg+2H2SO4→MgSO4+SO2+3H2O
Cu+2H2SO4→CuSO4+SO2+2H2O
Tho PT: ∑nH2SO4=2∑nSO2=0,2(mol)
→maxit=0,2.98=19,6(g)
Gọi nMg = a (mol); nCu = b (mol)
nSO2 = 2,24/22,4 = 0,1 (mol)
PTHH:
Mg + 2H2SO4 -> MgSO4 + SO2 + 2H2O
Mol: a ---> 2a ---> a ---> a ---> 2a
Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O
Mol: b ---> 2b ---> b ---> b ----> 2b
Ta có:
24a + 64b = 4,4
a + b = 0,1
=> a = b = 0,05 (mol)
mMg = 0,05 . 24 = 1,2 (g)
mCu = 0,05 . 64 = 3,2 (g)
%mMg = 1,2/4,4 = 27,27%
%mCu = 100% - 27,27% = 72,73%
mH2SO4 = (2 . 0,05 + 2 . 0,05) . 98 = 19,6 (g)
