a) 

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

             0,4-->0,2------->0,4

=> \(m_{H_2O}=0,4.18=7,2\left(g\right)\)

b) V_O2 = 0,2.22,4 = 4,48 (l)

c)

C1: m_O2 = 0,2.32 = 6,4 (g)

C2:

Theo ĐLBTKL: \(m_{O_2}=m_{H_2O}-m_{H_2}=7,2-0,4.2=6,4\left(g\right)\)

a)2H₂ + O₂ --t^o--> 2H₂O

b) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

            0,5-->0,25------>0,5

=> \(m_{H_2O}=0,5.18=9\left(g\right)\)

c) V_O2 = 0,25.22,4 = 5,6 (l)

=> V_kk = 5,6 : 20% = 28 (l)

\(n_{H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)

PTHH: \(2H_2+O_2\xrightarrow[]{t^\circ}2H_2O\)

       0,125 → 0,0625 → 0,125

a) \(V_{O_2}=0,0625\cdot22,4=1,4\left(l\right)\)

b) \(m_{H_2O}=0,125\cdot18=2,25\left(g\right)\)

\(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

C₂H₄ + 3O₂ ----t^o---> 2CO₂ + 2H₂O

0,4        1,2                                0,8

\(m_{H_2O}=0,8.18=14,4\left(g\right)\)

\(V_{kk}=5V_{O_2}=5.1,2.22,4=134,4\left(l\right)\)

a) nFe=0,45mol

PTHH: 2Fe+O2=>2FeO

              0,45->0,225

=> VO2 cần dùng =0,225.22,4=5,04 lít

b)2KClO3=>2KCl+3O2

      0,15<---------------0,225

=> mKClO3=0,15.122,5=18,375g

A.

Số mol của Fe: n=\(\frac{m}{M}\) =\(\frac{25,2}{56}\) = 0.45 (mol)

2Fe + O₂ --t⁰-> 2FeO

Theo PT 2 : 1 : 2

Theo bài ra 0.45 : 0.225 : 0.45 (mol)

Thể tích Oxi tham gia phảm ứng: V = n . 22,4 = 5.04 ( lít )

B.

Ta có: 2KClO₃ -t⁰-> 2KCl + 3O₂

Theo PT 2 : 2 : 3

Theo bài ra 0,15 : 0,15 : 0,225 (mol)

Khối lượng KClO₃ : m = n.M = 0.15 . 122,5 = 18,375 (g)

mMg = 3.6/24 = 0.15 (mol) 

2Mg + O2 -to-> 2MgO 

0.15__0.075____0.15 

mMgO= 0.15*40 = 6 (g) 

VO2  = 0.075*22.4 = 1.68 (l) 

2KClO3 -to-> 2KCl + 3O2 

0.05_______________0.075 

mKClO3 = 0.05*122.5 = 6.125 (g) 

PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

a+b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{MgO}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{MgO}=0,15\cdot40=6\left(g\right)\end{matrix}\right.\)

c) PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)

Theo PTHH: \(n_{KClO_3}=0,05\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=0,05\cdot122,5=6,125\left(g\right)\)

a, PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

Ta có: \(n_{C_2H_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Theo PT: \(n_{O_2}=3n_{C_2H_4}=0,75\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,75.2,24=16,8\left(l\right)\)

\(\Rightarrow V_{kk}=16,8.5=84\left(l\right)\)

b, Theo PT: \(n_{CO_2}=2n_{C_2H_4}=0,5\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{\downarrow}=m_{CaCO_3}=0,5.100=50\left(g\right)\)

\(m_{Ca\left(OH\right)_2}=0,5.74=37\left(g\right)\)

\(\Rightarrow m_{ddCa\left(OH\right)_2}=\dfrac{37.100}{2}=1850\left(g\right)\)

Bạn tham khảo nhé!

Đáp án:

VO2=22,4 (l); VCO2=11,2 (l)��2=22,4 (�); ���2=11,2 (�)

Giải thích các bước giải:

Ta có:

a, \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

Theo PT: \(n_{CO_2}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{15}\left(mol\right)\Rightarrow V_{CO_2}=\dfrac{2}{15}.22,4=\dfrac{224}{75}\left(l\right)\)

b, \(n_{C_2H_6O\left(LT\right)}=\dfrac{1}{3}n_{O_2}=\dfrac{1}{15}\left(mol\right)\)

Mà: H = 90%

\(\Rightarrow n_{C_2H_6O\left(TT\right)}=\dfrac{\dfrac{1}{15}}{90\%}=\dfrac{2}{27}\left(mol\right)\)

\(\Rightarrow m_{C_2H_6O}=\dfrac{2}{27}.46=\dfrac{92}{27}\left(g\right)\)