\(n_{CaCO_3}=\dfrac{7,5}{100}=0,075\left(mol\right)\)

=> n_C = 0,075 (mol)

Có \(n_{CO_2}=n_C=0,075\left(mol\right)\)

=> \(n_{H_2O}=\dfrac{4,2-0,075.44}{18}=0,05\left(mol\right)\)

=> n_H = 0,1 (mol)

\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Bảo toàn O: \(n_{O\left(A\right)}=0,075.2+0,05-0,1.2=0\left(mol\right)\)

=> A chứa C, H

m_A = m_C + m_H = 0,075.12 + 0,1.1 = 1 (g)

\(m_{tăng}=m_{H_2O}+m_{CO_2}=4,2\left(g\right)\\ n_{CaCO_3}=\dfrac{7,5}{100}=0,075\left(mol\right)\)

PTHH: Ca(OH)₂ + CO₂ ---> CaCO₃ + H₂O

                            0,075        0,075

\(\rightarrow m_{CO_2}=0,075.44=3,3\left(g\right)\\ \rightarrow m_{H_2O}=4,2-3,3=0,9\left(g\right)\\ \rightarrow n_{H_2O}=\dfrac{0,9}{18}=0,05\left(mol\right)\\ \rightarrow n_{O\left(sau.pư\right)}=0,05+0,075.2=0,1\left(mol\right)\\ n_{O\left(trong.O_2\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}n_C=0,075\left(mol\right)\\n_H=0,05.2=0,1\left(mol\right)\\n_O=0,1-0,1=0\left(mol\right)\end{matrix}\right.\)

=> m_A = 0,075.12 + 0,1.1 + 0 = 1 (g)

a)

CTHH: Fe_xO_y

\(n_{Fe_xO_y}=\dfrac{16}{56x+16y}\left(mol\right)\)

PTHH: Fe_xO_y + yCO --t^o--> xFe + yCO₂

      \(\dfrac{16}{56x+16y}\)--------->\(\dfrac{16x}{56x+16y}\)

=> \(\dfrac{16x}{56x+16y}.56=16-4,8=11,2\)

=> \(\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow Fe_2O_3\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 3CO --t^o--> 2Fe + 3CO₂

              0,1------>0,3--------------->0,3

             Ca(OH)₂ + CO₂ --> CaCO₃ + H₂O

                                0,3----->0,3

=> \(m_{CaCO_3}=0,3.100=30\left(g\right)\)

b) n_{CO (thực tế)} = 0,3.110% = 0,33(mol)

=> V_CO = 0,33.22,4 = 7,392(l)

nMg = 5,76/24 = 0,24 (mol)

PTHH: 2Mg + O2 -> (t°) 2MgO

nMgO = 0,24 (mol)

mMgO = 0,24 . 40 = 9,6 (g)

nMg = 5,76 : 24 = 0,24 ( mol ) 
pthh : 2Mg+ O2 -t--> 2MgO
          0,24->0,12-->0,24 (mol) 
=> m = mMgO = 0,24 . 40 = 9,6 (g) 

Bài 1 : 

a. \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\)

b. PTHH : 4Al + 3O₂ -t^o> 2Al₂O₃

                 0,4       0,3       0,2

Xét tỉ lệ : \(\dfrac{0,4}{4}< \dfrac{0,5}{3}\) => Al đủ , O₂ dư

\(m_{O_2\left(dư\right)}=\left(0,5-0,3\right).32=6,4\left(g\right)\)

c. \(m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

Bài 2:

nCH4 = 11.2/22.4 = 0.5 (mol) 

CH4 + 2O2 -to-> CO2 + 2H2O 

0.5____________0.5 

CO2 + Ca(OH)2 => CaCO3 + H2O 

0.5_______________0.5 

mCaCO3 = 0.5*100 = 50 (g) 

\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 +2H_2O\\ CO_2 +C a(OH)_2 \to CaCO_3 + H_2O\\ n_{CaCO_3} = n_{CO_2} = n_{CH_4} = \dfrac{11,2}{22,4} = 0,5(mol)\\ m = 0,5.100 = 50(gam)\)

a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

b, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,4}{5}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\)

c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

d, \(m_{P_2O_5}=14,2.80\%=11,36\left(g\right)\)

Giúp mik vs T-T