Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,2----------->0,2----->0,3
=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b) \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
c)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
0,3<----------------0,3
=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
\(a,n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2--------------->0,2------->0,3
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\\ b,m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
c, PTHH:
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,2<------------------0,2
\(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)
c, \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\Rightarrow m_{AlCl_3}=0,3.133,5=40,05\left(g\right)\)
d, \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Có: \(\dfrac{0,2}{1}>\dfrac{0,45}{3}\) → Fe2O3 dư.
\(n_{Fe}=\dfrac{2}{3}n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,1 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
\(m_{AlCl_3}=0,1.133,5=13,35g\)
d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,15 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
2Al + 6HCl → 2AlCl3 + 3H2
nAl = \(\dfrac{3,375}{27}\)= 0,125 mol
a) Theo tỉ lệ phản ứng => nH2 = \(\dfrac{3}{2}\)nAl = 0,1875 mol
<=> V H2 = 0,1875.22,4 = 4,2 lít
b) nAlCl3 = nAl = 0,125 mol
=> mAlCl3 = 0,125 . 133,5 = 16,6875 gam
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,2 0,3 ( mol )
\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)
\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,2.133,5=26,7g\)
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)
b, Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)
\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)
0,2<--0,6<----------0,2<------0,3 (mol)
\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(m_{HCl}=n\cdot M=0,6\cdot\left(1+35,5\right)=21,9\left(g\right)\)
\(m_{AlCl_3}=n\cdot M=0,2\cdot\left(27+35,5\cdot3\right)=26,7\left(g\right)\)
a, PT: 2Al+6HCl→2AlCl3+3H2
Ta có: nH2=6,7222,4=0,3(mol)
Theo PT: nHCl=2nH2=0,6(mol)
⇒mHCl=0,6.36,5=21,9(g)
b, Theo PT: nAl=23nH2=0,2(mol)
⇒mAl=0,2.27=5,4(g)
nAl = 8,1 /27 = 0,3mol
2Al + 6HCl => 2AlCl3 + 3H2
0,3--------------->0,3------> 0,45
=> VH2 = 0,45.22,4 = 10,08 (l)
mAlCl3 = 0,3. 133,5 = 40,05 (g)
cảm ơn nhé