Câu hỏi của Phù Thủy

Question

Tính khối lượng khí oxi cần dùng để đốt cháy hết hỗn hợp:
0,5 mol sắt; 1,25 mol nhôm; 1,5 mol kẽm.
3,1 gam P; 6,4 gam S; 3,6 gam C.
1,6 gam CH4; 2,8 gam CO; 0,58 gam C4H10.

Nguyễn Trần Thành Đạt · Accepted Answer

Mình cho a,b,c,d,... nha!

a) 0,5 mol Fe.

PTHH: 3Fe + 2O2 -to-> Fe3O4

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{2.n_{Fe}}{3}=\frac{2.0,5}{3}\approx0,333\left(mol\right)$

=> $m_{O_2}=0,333.32=10,656\left(g\right)$

b) 1,25 mol nhôm

PTHH: 4Al + 3O2 -to-> 2Al2O3

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{3.1,25}{4}=0,9375\left(mol\right)$

=> $m_{O_2}=32.0,9375=30\left(g\right)$

c) 1,5 mol Zn

2Zn + O2 -to-> 2ZnO

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{1,5}{2}=0,75\left(mol\right)$

=> $m_{O_2}=0,75.32=24\left(g\right)$Câu trả lời: Mình cho a,b,c,d,... nha!

a) 0,5 mol Fe.

PTHH: 3Fe + 2O2 -to-> Fe3O4

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{2.n_{Fe}}{3}=\frac{2.0,5}{3}\approx0,333\left(mol\right)$

=> $m_{O_2}=0,333.32=10,656\left(g\right)$

b) 1,25 mol nhôm

PTHH: 4Al + 3O2 -to-> 2Al2O3

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{3.1,25}{4}=0,9375\left(mol\right)$

=> $m_{O_2}=32.0,9375=30\left(g\right)$

c) 1,5 mol Zn

2Zn + O2 -to-> 2ZnO

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{1,5}{2}=0,75\left(mol\right)$

=> $m_{O_2}=0,75.32=24\left(g\right)$

Nguyễn Trần Thành Đạt · Answer

d) Ta có:

$n_P=\frac{3,1}{31}=0,1\left(mol\right)$

PTHH: 4P + 5O2 -to-> 2P2O5

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{5.0,1}{4}=0,125\left(mol\right)$

=> $m_{O_2}=0,125.32=4\left(g\right)$

e) Ta có:

$n_S=\frac{6,4}{32}=0,2\left(mol\right)$

PTHH: S + O2 -to-> SO2

Theo PTHH và đề bài, ta có:

$n_{O_2}=n_S=0,2\left(mol\right)\ =>m_{O_2}=0,2.32=6,4\left(g\right)$

f) Ta có:

$n_C=\frac{3,6}{12}=0,3\left(mol\right)$

PTHH: C + O2 -to-> CO2

Theo PTHH và đề bài, ta có:

$n_{O_2}=n_C=0,3\left(mol\right)$

=> $m_{O_2}=32.0,3=9,6\left(g\right)$Câu trả lời: d) Ta có:

$n_P=\frac{3,1}{31}=0,1\left(mol\right)$

PTHH: 4P + 5O2 -to-> 2P2O5

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{5.0,1}{4}=0,125\left(mol\right)$

=> $m_{O_2}=0,125.32=4\left(g\right)$

e) Ta có:

$n_S=\frac{6,4}{32}=0,2\left(mol\right)$

PTHH: S + O2 -to-> SO2

Theo PTHH và đề bài, ta có:

$n_{O_2}=n_S=0,2\left(mol\right)\ =>m_{O_2}=0,2.32=6,4\left(g\right)$

f) Ta có:

$n_C=\frac{3,6}{12}=0,3\left(mol\right)$

PTHH: C + O2 -to-> CO2

Theo PTHH và đề bài, ta có:

$n_{O_2}=n_C=0,3\left(mol\right)$

=> $m_{O_2}=32.0,3=9,6\left(g\right)$

Nguyễn Trần Thành Đạt · Answer

g) Ta có:

$n_{CH_4}=\frac{1,6}{16}=0,1\left(mol\right)$

PTHH: CH4 + 2O2 -> CO2 + 2H2O

Theo PTHH và đề bài, ta có:

$n_{O_2}=2.n_{CH_4}=2.0,1=0,2\left(mol\right)\
=>m_{O_2}=32.0,2=6,4\left(g\right)$

h) Ta có:

$n_{CO}=\frac{2,8}{28}=0,1\left(mol\right)$

PTHH: 2CO + O2 -to-> 2CO2

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{0,1}{2}=0,05\left(mol\right)\
=>m_{O_2}=0,05.32=1,6\left(g\right)$

i) Ta có:

$n_{C_4H_{10}}=\frac{0,58}{58}=0,01\left(mol\right)$

PTHH: 2C4H10 + 13 O2 -to-> 8 CO2 +10  H2O

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{13.0,01}{2}=0,065\left(mol\right)$

=>$m_{O_2}=32.0,065=2,08\left(g\right)$Câu trả lời: g) Ta có:

$n_{CH_4}=\frac{1,6}{16}=0,1\left(mol\right)$

PTHH: CH4 + 2O2 -> CO2 + 2H2O

Theo PTHH và đề bài, ta có:

$n_{O_2}=2.n_{CH_4}=2.0,1=0,2\left(mol\right)\
=>m_{O_2}=32.0,2=6,4\left(g\right)$

h) Ta có:

$n_{CO}=\frac{2,8}{28}=0,1\left(mol\right)$

PTHH: 2CO + O2 -to-> 2CO2

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{0,1}{2}=0,05\left(mol\right)\
=>m_{O_2}=0,05.32=1,6\left(g\right)$

i) Ta có:

$n_{C_4H_{10}}=\frac{0,58}{58}=0,01\left(mol\right)$

PTHH: 2C4H10 + 13 O2 -to-> 8 CO2 +10  H2O

Theo PTHH và đề bài, ta có:

$n_{O_2}=\frac{13.0,01}{2}=0,065\left(mol\right)$

=>$m_{O_2}=32.0,065=2,08\left(g\right)$

CHƯƠNG IV: OXI - KHÔNG KHÍ

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