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a)
\(2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\)
b)
\(n_{H_2} = \dfrac{6.13,44}{22,4} = 3,6(mol)\)
Theo PTHH :
\(n_{Al} = \dfrac{2}{3}n_{H_2} = 2,4(mol)\\ \Rightarrow m_{Al} = 2,4.27 = 64,8(gam)\)
c)
\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)
Theo PT trên :
\(n_{O_2} = \dfrac{3}{4}n_{Al} = 1,8(mol)\\ \Rightarrow V_{O_2} = 1,8.22,4 = 40,32(lít)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,15\left(mol\right)=n_{H_2}\\n_{Al_2\left(SO_4\right)_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)
a) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
______0,1--->0,15-------->0,05------->0,15
=> mH2SO4 = 0,15.98 = 14,7 (g)
b) VH2 = 0,15.22,4 = 3,36 (l)
c) mAl2(SO4)3 = 0,05.342 = 17,1 (g)
V H2=13,44 lít nha
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có:\(n_{H2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(n_{Al}=\frac{2}{3}n_{H2}=0,4\left(mol\right)\)
\(\Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)
\(4Al+3O_2\rightarrow2Al_2O_3\)
Ta có: \(n_{O2}=\frac{3}{4}n_{Al}=\frac{3}{4}.0,4=0,3\left(mol\right)\Rightarrow V_{O2}=0,3.22,4=6,72\left(l\right)\)
a: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b: \(n_{H2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(\Leftrightarrow n_{Al}=0.1\left(mol\right)\)
\(m_{Al}=n_{Al}\cdot M_{Al}=0.1\cdot27=2.7\left(g\right)\)
số mol H2 là:\(n=\frac{613.44}{22.4}\approx27.39\left(mol\right)\)
PTHH\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ m_{Al}=\left(27.39\cdot\frac{2}{3}\right)\cdot27=493.02\left(g\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
18,25______________________27,39
\(n_{H2}=27,39\left(mol\right)\)
\(\Rightarrow m=492,75\left(g\right)\)