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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al_2S_3}=\dfrac{25,5}{150}=0,17\left(mol\right)\)
PTHH: 2Al + 3S --to--> Al2S3
0,34<----------0,17
=> \(H\%=\dfrac{0,34.27}{10,8}.100\%=85\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al_2S_3}=\dfrac{25.5}{150}=0.17\left(mol\right)\)
\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)
\(2Al+3S\underrightarrow{^{t^0}}Al_2S_3\)
\(0.34...........0.17\)
\(H\%=\dfrac{0.34}{0.4}\cdot100\%=85\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nAl=10/27(mol)
ta ccó pthh: 2Al+3S->Al2S3( nhiệt dộ cao)
theo ptth=> nAl2S3(lý thuyết)=1/2.nAl=\(\dfrac{1}{2}.\dfrac{10}{27}\)=\(\dfrac{5}{27}\)(mol)
=> mAl2S3(lý thuyết)=\(\dfrac{5}{27}.150=\dfrac{250}{9}\)(g)
=>H=\(\dfrac{mAL2S3\left(thucte\right)}{mAL2S3\left(lythuyet\right)}.100\%=\dfrac{25,5}{\dfrac{250}{9}}=91,8\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PT: \(2Al+3S\underrightarrow{t^o}Al_2S_3\)
Theo PT: \(n_{Al_2S_3\left(LT\right)}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al_2S_3\left(LT\right)}=0,2.150=30\left(g\right)\)
\(\Rightarrow H=\dfrac{25,5}{30}.100\%=85\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al_2S_3\left(TT\right)}=\dfrac{25,5}{150}=0,17\left(mol\right)\\ n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ 2Al+3S\rightarrow\left(t^o\right)Al_2S_3\\ Ta,có:n_{Al_2S_3\left(LT\right)}=\dfrac{1}{2}n_{Al}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ H=\dfrac{0,17}{0,2}.100\%=85\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: 2Al + 3S ===> Al2S3
=> nAl = 8,1 / 27 = 0,3 mol
=> nS = 9,6 / 32 = 0,3 mol
Lập tỉ lệ ===> Al dư, S hết
=> nAl(pứ) = 0,2 mol
=> mAl(pứ) = 0,2 x 27 = 5,4 gam
Áp dụng định luật bảo toàn khối lượng
=> mAl2S3 = 5,4 + 9,6 = 15 gam
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3 :
a) \(Theo-\text{đ}\text{è}-b\text{ài}-ta-c\text{ó}:\left\{{}\begin{matrix}nCu=\dfrac{51,2}{64}=0,8\left(mol\right)\\nCuO=\dfrac{48}{80}=0,6\left(mol\right)\end{matrix}\right.\)
Ta có PTHH :
2Cu + O2 -t0\(\rightarrow\) 2CuO
Theo PTHH ta có : nCu = \(\dfrac{0,8}{2}mol>nCuO=\dfrac{0,6}{2}mol\) => nCu dư
b) Ta có : H = \(\dfrac{n\left(ch\text{ất}-thi\text{ếu}\right)}{n\left(ch\text{ất}-d\text{ư}\right)}.100\%=\dfrac{0,6}{0,8}.100\%=75\%\)
Vậy.......
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: nAl=5,4:27=0,2 mol
nS=6,4:32=0,2 mol
PTHH: 2Al + 3S \(\rightarrow\) Al2S3
ban đầu: 0,2 0,2 (mol)
phản ứng: 0,2 \(\leftarrow\) 0,2 (mol)
sau PƯ: 0 0 \(\frac{1}{15}\) (mol)
vậy sau phản ứng Al dư, S hết ( nhưng do cùng số mol nên Al hết)
mAL2S3= \(\frac{1}{15}.150=10\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2Al + 3S \(\underrightarrow{to}\) Al2S3
\(n_{Al_2S_3}=\frac{25,5}{150}=0,17\left(mol\right)\)
Theo PT: \(n_{Al}=2n_{Al_2S_3}=2\times0,17=0,34\left(mol\right)\)
\(\Rightarrow m_{Al}=0,34\times27=9,18\left(g\right)\)
\(\Rightarrow H\%=\frac{9,18}{10,8}\times100\%=85\%\)
2Al + 3S---t0--> Al2S3
Ta có nAl=10,8/27=0,4
nAl2S3=25,5/150=0,17
=> nAl đã PỨ= 2nAl2O3=0,34
=> H%=0,34.100/0,4=85%
2Al + 3S -> Al2S3 (1)
nAl2S3=\(\dfrac{64}{375}\left(mol\right)\)
nAl=0,4(mol)
Từ 1:
nAl PƯ=2nAl2S3=\(\dfrac{128}{375}\left(mol\right)\)
H=\(\dfrac{128}{375}:0,4.100\%=85,3\%\)
mAl2S3=25,6 chia cho MAl2S3=150thanh thanh ngan