a)

\(4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\)

Sản phẩm : Điphotpho pentaoxit.

b)

\(n_P = \dfrac{6,2}{31} = 0,2(mol)\\ \Rightarrow n_{P_2O_5} = \dfrac{1}{2}n_P = 0,1(mol)\\ \Rightarrow m_{P_2O_5} = 0,1.142 = 14,2(gam)\)

c)

\(n_{O_2} = \dfrac{5}{4}n_P = 0,125(mol)\\ \Rightarrow V_{O_2} = 0,125.22,4 = 2,8(lít)\)

d)

\(V_{không\ khí} = \dfrac{2,8}{20\%} = 14(lít)\)

Giúp mình với mình đang cần gấp😇

\(n_{C_2H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ a,2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ b,n_{CO_2}=0,125.2=0,25\left(mol\right)\\ m_{CO_2}=0,25.44=11\left(g\right)\\ c,n_{O_2}=\dfrac{5}{2}.0,125=0,3125\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,3125.22,4=7\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=\dfrac{100}{20}.7=35\left(l\right)\)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

0,6       0,3                  0,6        ( mol )

\(m_{H_2O}=0,6.18=10,8g\)

\(V_{kk}=V_{O_2}.5=\left(0,3.22,4\right).5=33,6l\)

\(a) 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ b) n_{Cu}=\dfrac{12,7}{64} = \dfrac{127}{640}(mol)\\ \Rightarrow n_{O_2} = \dfrac{1}{2}n_{Cu} = \dfrac{127}{1280}(mol)\\ \Rightarrow m_{O_2} = \dfrac{127}{1280}.32 = 3,175(gam)\\ c) V_{không\ khí} = 5V_{O_2} = 5.\dfrac{127}{1280}.22,4 = 11,1125(lít) \)

\(a,CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

 Vì n và V tỉ lệ thuận với nhau. Nên ta có:

\(V_{O_2}=2.V_{CH_4}=2.2,768=5,536\left(l\right)\)

\(b,V_{kk}=\dfrac{100}{21}.V_{O_2}=\dfrac{100}{21}.5,536=\dfrac{2768}{105}\left(l\right)\)

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\a, 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\b,n_{P_2O_5}=\dfrac{2}{5}.0,25=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=142.0,1=14,2\left(g\right)\\c,V_{kk\left(đktc\right)}=4.5,6=28\left(lít\right) \)

\(n_{H_2}\)=\(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH 2H₂ +O₂----t^o--->2H₂O

           0,2....0,1.................0,2

=>\(m_{H_2O}=0,2.18=3,6\left(g\right)\)

=>\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)

=>V_kk=2,24.5=11,2(l)

\(n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{H_2O} = n_{H_2} =0,2(mol) \Rightarrow m_{H_2O} = 0,2.18 = 3,6(gam)\\ n_{O_2} = \dfrac{1}{2}n_{H_2} = 0,1(mol)\\ \Rightarrow V_{O_2} = 0,1.22,4 = 2,24(lít)\\ \Rightarrow V_{không\ khí} = 5V_{O_2} = 2,24.5 = 11,2(lít) \)

a)\(2Mg + O_2 \xrightarrow{t^o} 2MgO\)

b)

\(n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)\)

Theo PTHH : 

\(n_{O_2} = \dfrac{1}{2}n_{Mg} = 0,05(mol)\\ \Rightarrow V_{O_2} = 0,05.22,4 = 1,12(lít)\)

c)

\(n_{MgO} = n_{Mg} = 0,1(mol)\\ \Rightarrow m_{MgO} = 0,1.40 = 4(gam)\)

d)

\(V_{không\ khí} = 5V_{O_2} = 1,12.5 = 5,6(lít)\)

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

C₂H₄ + 3O₂ ----t^o---> 2CO₂ + 2H₂O

0,4        1,2                                0,8

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

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