Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
4Al + 3O2 --to--> 2Al2O3
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
___________0,15<------0,1
=> mO2 = 0,15.32 = 4,8(g)
Bảo toàn KL: \(m_{Al}+m_{O_2}=m_{Al_2O_3}\)
\(\Rightarrow m_{O_2}=10,2-9=1,2(g)\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
a, Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=1,2\left(mol\right)\)
\(\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)
b, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)
nAl = \(\frac{13,5}{27}= 0,5\) mol
Pt: 4Al + 3O2 --to--> 2Al2O3
....0,5 mol------------> 0,25 mol
mAl2O3 sinh ra = 0,25 . 102 = 25,5 (g)
a) 4Al + 3O2 --to--> 2Al2O3
b) Theo ĐLBTKL: mAl + mO2 = mAl2O3 (1)
c) (1) => mAl = 10,2 - 4,8 = 5,4(g)
\(n_{O_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
a, \(n_{Al}=\dfrac{4}{3}n_{O_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)
b, \(n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)
a) Theo định luật bào toàn khối lượng , ta có :
mAl2O3 = mAl + mO2
b) Ta có :
mAl2O3 = 54 + 48 = 102 (g)
c) %mAl2O3 = \(\frac{102.100}{150}\%=68\%\)
a)
- PTHH: \(Al_2O_3\rightarrow Al+O_2\)
- Công thức về khối lượng: \(m_{Al_2O_3}=m_{Al}+m_{O_2}\)
b)
\(m_{Al_2O_3}=m_{Al}+m_{O_2}\)
hay \(m_{Al_2O_3}=54+48\)
\(\Rightarrow m_{Al_2O_3}=102\left(g\right)\)
c)
Phần trăm: \(m_{Al_2O_3}\) = \(m_{Al_2O_3}\) / m quặng boxit
\(\frac{150}{102}.100\%=1,5\%\)
câu c mk cũng hk chắc nha bạn!!!!!!!!!
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
______0,1->0,075-->0,05
a) VO2 = 0,075.22,4 = 1,68(l)
b) mAl2O3 = 0,05.102 = 5,1 (g)
\(n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ a.PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
4 3 2
0,4 0,3 0,2
\(m_{Al_2O_3}=n.M=0,2.\left(27.2+16.3\right)=20,4\left(g\right)\\ c.V_{O_2}=n.24,79=0,3.24,79=7,437\left(l\right)\)
\(d.n_{O_2}=\dfrac{m}{M}=\dfrac{12,8}{\left(16.2\right)}=0,4\left(mol\right)\\ PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
4 3 2
0,53 0,4 0,27
Tỉ lệ: \(\dfrac{0,53}{4}< \dfrac{0,4}{3}< \dfrac{0,27}{2}\Rightarrow Al_2O_3\) dư và dư \(m_{Al_2O_3}=n.M=0,27.\left(27.2+16.3\right)=27,54\left(g\right).\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\Rightarrow m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)
c, \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)
d, \(n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,4}{4}< \dfrac{0,4}{3}\), ta được O2 dư.
\(\Rightarrow n_{O_2\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\Rightarrow m_{O_2\left(dư\right)}=0,1.32=3,2\left(g\right)\)
a/ nAl= 54/27= 2(mol)
nO2=48/32=1,5(mol)
PTHH: 2 Al2O3 -to-> 4 Al +3 O2
Ta có: 2/4 = 1,5/3
=> P.ứ hết
=> nAl2O3= 1/2. nAl=1/2. 2=1(mol)
=> mAl2O3=1.102=102(g)
b) %mAl2O3= (102/127,5).100= 80%
\(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1mol\\ 4Al+3O_2\xrightarrow[]{t^0}2Al_2O_3\\ n_{O_2}=\dfrac{0,1.3}{2}=0,15mol\\ m_{O_2}=0,15.32=4,8g\)