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\(n_{Al}=\dfrac{2,5}{27}=\dfrac{25}{270}=\dfrac{5}{54}\left(mol\right)\\ n_{H_2SO_4}=0,5\left(mol\right)\\ a,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ b,Vì:\dfrac{\dfrac{5}{54}}{2}< \dfrac{0,5}{4}\Rightarrow H_2SO_4dư\\ b,n_{H_2SO_4\left(dư\right)}=0,5-\dfrac{3}{2}.\dfrac{5}{54}=\dfrac{13}{36}\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=\dfrac{13}{36}.98=\dfrac{637}{18}\left(g\right)\\ c,n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.\dfrac{5}{54}=\dfrac{5}{36}\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=\dfrac{5}{36}.22,4=\dfrac{28}{9}\left(l\right)\)
2Al+3H2SO4->Al2(SO4)3+3H2
0,2-----0,3-------0,1------------0,3
n Al=\(\dfrac{5,4}{27}\)=0,2 mol
n H2SO4= \(\dfrac{30}{98}\)=0,306 mol
=>H2SO4 còn dư
=>VH2=0,3.22,4=6,72l
=>m Al2(SO4)3=0,1.342=34,2g
=>m H2SO4 dư=0,006.98=0,588g
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)
\(n_{H_2SO_4}=\dfrac{m_{H_2SO_4}}{M_{H_2SO_4}}=\dfrac{30}{98}=\dfrac{15}{49}mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
2 3 1 3 ( mol )
0,2 15/49 ( mol )
Ta có: \(\dfrac{0,2}{2}< \dfrac{15}{49}:3\)
=> Chất còn dư là \(H_2SO_4\)
\(V_{H_2}=n_{H_2}.22,4=\left(\dfrac{0,2.3}{2}\right).22,4=6,72l\)
\(m_{Al_2\left(SO_4\right)_3}=n_{Al_2\left(SO_4\right)_3}.M_{Al_2\left(SO_4\right)_3}=\left(\dfrac{0,2.1}{2}\right).342=34,2g\)
\(m_{H_2SO_4\left(du\right)}=n_{H_2SO_4\left(du\right)}.M_{H_2SO_4}=\left(\dfrac{15}{49}-\dfrac{0,2.3}{2}\right).98=0,6g\)
a: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b: \(n_{Al}=\dfrac{2.5}{27}< \dfrac{1}{4}\)
=>H2SO4 dư, Al đủ
\(m_{H_2SO_4}=0.25\cdot98=24.5\left(g\right)\)
c: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{2.5}{54}=\dfrac{5}{108}\left(mol\right)\)
\(\Leftrightarrow n_{H_2}=\dfrac{5}{36}\left(mol\right)\)
\(V_{H_2}=\dfrac{5}{36}\cdot22.4=\dfrac{28}{9}\left(lít\right)\)
Mình thấy bạn Thịnh tính lượng dư sai
Đây là bài mình từng làm, bạn tham khảo nhé!
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
Làm gộp các phần còn lại
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\)
a) \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,25 0,5 0,5 0,5
Xét tỉ lệ : \(\dfrac{0,3}{1}>\dfrac{0,5}{2}\) => Zn dư , HCl đủ
b) \(m_{Zn\left(dư\right)}=\left(0,3-0,25\right).65=3,25\left(g\right)\)
c) \(m_{ZnCl_2}=0,25.136=34\left(g\right)\)
\(V_{H_2}=0,25.22,4=5,6\left(l\right)\)
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ a,Zn+2HCl\rightarrow ZnCl_2+H_2\\b, Vì:\dfrac{0,5}{2}< \dfrac{0,3}{1}\Rightarrow Zndư\\ n_{Zn\left(dư\right)}=0,3-\dfrac{0,5}{2}=0,05\left(mol\right)\\ \Rightarrow m_{Zn\left(dư\right)}=0,05.65=3,25\left(g\right)\\ c,n_{ZnCl_2}=n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,25.136=34\left(g\right)\\ V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PTHH : Fe + 2HCl -> FeCl2 + H2
0,2 0,4 0,2
Xét tỉ lệ \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => Fe đủ , HCl dư
\(m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65\left(g\right)\)
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49}{98}=0.5\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.2........0.3.................................0.3\)
\(m_{H_2SO_4\left(dư\right)}=\left(0.5-0.3\right)\cdot98=19.6\left(g\right)\)
\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)
a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\\n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,5}{1}>\dfrac{0,1}{2}\) \(\Rightarrow\) HCl phản ứng hết, Zn còn dư
\(\Rightarrow n_{Zn\left(dư\right)}=0,5-0,05=0,45\left(mol\right)\) \(\Rightarrow m_{Zn\left(dư\right)}=0,45\cdot65=29,25\left(g\right)\)
c+d) Theo PTHH: \(n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05mol\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,05\cdot136=6,8\left(g\right)\\V_{H_2}=0,05\cdot22,4=1,12\left(l\right)\end{matrix}\right.\)
Chúc bạn học tốt!