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\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\ a,m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\\ n_{H_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,n_{HCl}=\dfrac{6}{2}.0,1=0,3\left(mol\right)\\ c,C_{MddHCl}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
c, Ta có: m dd sau pư = 8,1 + 200 - 0,45.2 = 207,2 (g)
Theo PT: \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,3.133,5}{207,2}.100\%\approx19,33\%\)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
Ta có: \(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
a, Theo PT: \(n_{H_2\left(LT\right)}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\)
\(\Rightarrow V_{H_2\left(LT\right)}=0,15.24,79=3,7185\left(l\right)\)
\(\Rightarrow H=\dfrac{V_{H_2\left(TT\right)}}{V_{H_2\left(LT\right)}}.100\%=\dfrac{2,479}{3,7185}.100\%\approx66,67\%\)
b, \(n_{HCl}=\dfrac{6}{2}n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{n_{HCl}}{C_{M_{HCl}}}=\dfrac{0,3}{1,5}=0,2\left(l\right)\)
c, \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
a.
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(n_{HCl}=\dfrac{29.2}{36.5}=0.8\left(mol\right)\)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Lập tỉ lệ :
\(\dfrac{0.2}{2}< \dfrac{0.8}{6}\rightarrow HCldư\)
Khi đó :
\(n_{AlCl_3}=0.2\left(mol\right),n_{H_2}=0.2\cdot\dfrac{3}{2}=0.3\left(mol\right)\)
\(n_{HCl\left(dư\right)}=0.8-0.2\cdot3=0.2\left(mol\right)\)
\(m_{HCl\left(dư\right)}=0.2\cdot36.5=7.3\left(g\right)\)
\(b.\)
\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)
\(c.\)
\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,2 0,2 0,3
\(V_{H_2}=n.22,4=6,72\left(l\right)\)
\(m_{AlCl_3}=n.M=0,2.133,5=26,7\left(g\right)\)
18,25 là số gam của dd mà sao tính đc công thức đấy , dd tính theo công thức n/V thôi chứ .
a:
\(n_{Al}=\dfrac{5.4}{27}=0,2\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,2 0,6 0,2 0,3
\(m_{AlCl_3}=0,2\left(27+35,5\cdot3\right)=26,7\left(g\right)\)
b: \(100ml=0,1\left(lít\right)\)
\(C_{M\left(HCl\right)}=\dfrac{0.6}{1}=0,6\left(\dfrac{mol}{lít}\right)\)
c: \(n_{H_2}=3\cdot\dfrac{0.2}{2}=0,3\left(mol\right)\)
=>\(V_{H_2}=0,3\cdot22,4=6,72\left(lít\right)\)