\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(PTHH:\)
\(M_g+2CH_3COOH\rightarrow M_g\left(CH_3COO\right)_2+H_2\)
\(0,2\rightarrow0,4\) \(\rightarrow0,2\) \(\rightarrow0,2\left(mol\right)\)
\(a,\) \(C_{MCH_3COOH}=\dfrac{0,4}{0,2}=2\left(M\right)\)
\(b,\) \(m_{Mg\left(CH_3COO\right)_2}=142.0,2=28,4\left(g\right)\)
\(c,\) \(sửa\) \(:\) \(thu\) \(được\) \(9,2g\) \(este\)
\(n_{CH_3COOC_2H_5}=\dfrac{9,2}{88}=0,1\left(mol\right)\)
\(CH_3COOH+C_2H_5OH\rightarrow CH_3COOC_2H_5+H_2O\)
\(thực\) \(ra:\) \(0,2\) \(0,1\) \(\left(mol\right)\)
\(k\)/\(thức\) \(:\) \(0,1\) \(0,1\) \(\left(mol\right)\)
\(H=\dfrac{0,1}{0,2}.100=50\%\)
nMg=4,824=0,2(mol)nMg=4,824=0,2(mol)
PTHH:
Mg+2CH3COOH→Mg(CH3COO)2+H2Mg+2CH3COOH→Mg(CH3COO)2+H2
0,2→0,40,2→0,4 →0,2→0,2 →0,2(mol)→0,2(mol)
a,a,
CMCH3COOH=0,40,2=2(M)CMCH3COOH=0,40,2=2(M)
b,b,
mMg(CH3COO)2=142.0,2=28,4(g)mMg(CH3COO)2=142.0,2=28,4(g)
c,c,
(Phản ứng với 9,2(g)9,2(g) ... gì ?)
đây là lần thứ 3 bn hỏi bài này rồi đấy :)
