Bài 10:

$a)PTHH:CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O$

$n_{CH_4}=\dfrac{17,92}{22,4}=0,8(mol)$

Theo PT: $n_{O_2}=n_{H_2O}=1,6(mol);n_{CO_2}=0,8(mol)$

$b)m_{H_2O}=1,6.18=28,8(g)$

$c)V_{CO_2}=0,8.22,4=17,92(lít)$

$d)V_{O_2}=1,6.22,4=35,84(lít)$

$\Rightarrow V_{kk}=35,84.5=179,2(lít)$

Bài 11:

$a)PTHH:C+O_2\xrightarrow{t^o}CO_2$

$b)$ Theo PT: $n_C=n_{CO_2}=n_{O_2}=\dfrac{6}{12}=0,5(mol)$

$\Rightarrow V_{CO_2}=0,5.22,4=11,2(lít)$

$\Rightarrow V_{kk}=0,5.22,4.5=56(lít)$

Bài 12:

$a)PTHH:Fe+S\xrightarrow{t^o}FeS$

$n_{Fe}=\dfrac{5,6}{56}=0,1(mol)$

Theo PT: $n_S=n_{FeS}=n_{Fe}=0,1(mol)$

$b)m_{FeS}=0,1.88=8,8(g)$

$m_S=0,1.32=3,2(g)$

Đáp án

BÀi 10:

nCH4=17,92/22,4=0,8(mol)

a)PTHH:CH4+2O2→(t°)CO2+2H2O

Theo pt:1mol 2mol    1 mol 2mol

Theo đề: 0,8mol 1,6mol 0,8mol 1,6mol

b)mH2O=1,6.18=28,8(g)

c)VCO2=0,8.22,4=17,92(l)

d)VO2=1,6.22,4=35,84(l)

→Vkk=5VO2=5.35,84=179,2(l)

Bài 11

a) nC= 0.5 mol

C + O2 -to-> CO2

0.5   0.5          0.5

b)VCO2 = 0.5*22.4 = 11.2 (l)

VO2 = 0.5*22.4 = 11.2 (l)

VKK = 11.2*5 = 56 (l)

Bài 12

 a ) Fe+  S - t độ→  FeS

      0.1    0.1  0.1

b) n Fe = 5.6/ 56 = 0.1 mol

mFeS = 0.1 x  88 = 8.8 g

m S = 0.1 x 32 = 3.2 g