a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b) \(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{30}{98}=0,31\left(mol\right)\)
Lập tỉ lệ :
\(\dfrac{n_{Al}}{2}=\dfrac{0,2}{2}=0,1\)
\(\dfrac{n_{H_2SO_4}}{3}=\dfrac{0,31}{3}=0,103\)
Ta thấy : \(\dfrac{n_{H_2SO_4}}{3}>\dfrac{n_{Al}}{2}\left(0,103>0,1\right)\)
\(\Rightarrow\) H2SO4 dư
b) \(n_{H_2}=n_{Al}\cdot\dfrac{3}{2}=0,2\cdot\dfrac{3}{2}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=n\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\)
c) \(n_{H_2SO_4pu}=n_{Al}\cdot\dfrac{3}{2}=0,2\cdot\dfrac{3}{2}=0,3\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4du}=n_{H_2SO_3bđ}-n_{H_2SO_4pu}=0,31-0,3=0,01\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4du}=n\cdot M=0,01\cdot98=0,98\left(g\right)\)
\(n_{Al_2\left(SO_4\right)_3}=n_{Al}\cdot\dfrac{1}{2}=0,2\cdot\dfrac{1}{2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=n\cdot M=0,1\cdot342=34,2\left(g\right)\)
\(m_{H_2}=0,3\cdot2=0,6\left(g\right)\)
a) PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
Ta có : nAl = \(\dfrac{5,4}{27}\) = 0,2 mol
\(n_{H_2SO_4}\) = \(\dfrac{30}{98}\) = 0,306 mol
Vì \(\dfrac{0,2}{2}< \dfrac{0,306}{3}\) => H2SO4 dư
b) Theo PT: \(n_{H_2}\) = 0,3 mol
=> \(V_{H_2}\) = 0,3 x 22,4 = 6,72l
c) \(m_{H_2SO_4}\) dư = \(\left(0,306-0,3\right)\times98\) = 0,588g
\(m_{Al_2\left(SO_4\right)_3}\) = \(0,1\times342\) = 34,2g
\(m_{H_2}\) = 0,3 x 2 = 0, 6g
\(2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)
nAl= \(\dfrac{5,4}{27}=0,2mol\)
\(n_{H2SO4_{ }}=\dfrac{30}{98}=0,306mol\)
Ta có \(\dfrac{0,2}{2}< \dfrac{0,306}{3}\) nên H2SO4 sẽ dư
nH2SO4 dư = 0,306- (0,2/2).3=0,006mol
Ta có nH2=3/2nAl= 0,3mol
=> VH2 = 0,3.22,4=6,72 lít
mH2= 0,3.2=0,6 gam
mAl2(SO4)3 = 0,2.342=68,4 gam
mH2SO4 dư=0,006.98=0,588 gam
Ta có pthh
2Al +3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2
Theo đề bài ta có
nAl=\(\dfrac{5,4}{27}=0,2\left(mol\right)\)
nH2SO4=\(\dfrac{30}{98}\approx0,306\left(mol\right)\)
a, Theo pthh
nAl=\(\dfrac{0,2}{2}mol< nH2SO4=\dfrac{0,306}{3}mol\)
-> Số mol của H2SO4 dư (tính theo số mol của Al )
Vậy Chát dư sau phản ứng là H2SO4
b, Theo pthh
nH2=nH2SO4=3/2nAl=3/2.0,2=0,3(mol)
-> VH2(đktc)=0,3.22,4=6,72(l)
c, Theo pthh
nAl2(SO4)3=1/2nAl=1/2.0,2=0,1 mol
-> mAl2(SO4)3=0,1.342=34,2 g
mH2SO4(dư)=(0,306-0,3).98=0,588 g
a, PTHH: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{30}{98}=\dfrac{15}{49}\left(mol\right)\\ =>\dfrac{0,2}{2}< \dfrac{\dfrac{15}{49}}{3}\)
=> Al hết, H2SO4 dư nên tính theo \(n_{Al}\)
b, \(n_{H_2\left(đktc\right)}=\dfrac{3.0,2}{2}=0,3\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)
c, Các chất sau phản ứng gồm: \(\left\{{}\begin{matrix}H_2SO_4\left(dư\right)\\Al_2\left(SO_4\right)_3\\H_2\end{matrix}\right.\)
Ta có: \(n_{H_2SO_4\left(f.ứ\right)}=\dfrac{3.0,2}{2}=0,3\left(mol\right)\\ =>\left\{{}\begin{matrix}n_{H_2SO_4\left(dư\right)}=\dfrac{15}{49}-0,3\approx0,006\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\)
Vậy: => \(\left\{{}\begin{matrix}m_{H_2SO_4\left(dư\right)}\approx0,006.98\approx0,588\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\\m_{H_2}=2.0,3=0,6\left(g\right)\end{matrix}\right.\)