Sửa đề: 5,6 (g) → 5,4 (g)
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ:\(\dfrac{0,5}{1}>\dfrac{0,2}{1}\), ta được CuO dư.
Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,2\left(mol\right)\)
⇒ nCuO (dư) = 0,5 - 0,2 = 0,3 (mol)
⇒ m chất rắn = mCuO (dư) + mCu = 0,3.80 + 0,2.64 = 36,8 (g)
a) Sửa đề: Al -> Fe
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1-------------------------->0,1
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b) \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)
PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
Xét tỉ lệ: \(0,5>0,1\Rightarrow CuO\) dư
Theo PT: \(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m=0,4.80+0,1.64=38,4\left(g\right)\)
