a) 4P + 5O₂ --t^o--> 2P₂O₅

b) \(n_P=\dfrac{15,5}{31}=0,5\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

           0,5--------------->0,25

=> m_P2O5 = 0,25.142 = 35,5 (g)

c) 

PTHH: P₂O₅ + 3H₂O --> 2H₃PO₄

              0,25------------>0,5

=> \(C\%=\dfrac{0,5.98}{100}.100\%=49\%\)

4P+5O2-to>2P2O5

0,5-------------0,25

n P=0,5 mol

=>m P2O5=0,25.142=35,5g

c)P2O5+3H2O->2H3PO4

    0,25---------------------0,5

=>C%=\(\dfrac{0,5.98}{100}100=49g\)

Cảm ơn bạn nha

4P2+5)2 ---> 2P2O5

Lần lượt tính mol các chất theo tỉ lệ là đc

d) P2O5 + 3H2O ---> 2H3PO4

2K+2H2O->2KOH+H2

0,1-----0,1-----0,1---0,05

n K=0,1 mol

=>m KOH=0,1.56=5,6g

=>VH2=0,05.22,4=1,12l

2H2+O2-to>2H2O

0,05--------------0,05

n O2=0,2 mol

=>O2 dư

=>mH2O=0,05.18=0,9g

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:

a, \(n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH:

CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

0,5--->1------------->0,5

Ca(OH)₂ + CO₂ ---> CaCO₃ + H₂O

                   0,5----->0,5

b, \(V_{O_2}=1.22,4=22,4\left(l\right)\)

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

Câu 1:

a, \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

b, \(n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{CO_2}=2n_{C_2H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{CO_2}=0,6.44=26,4\left(g\right)\)

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

Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,6\left(mol\right)\Rightarrow m_{CaCO_3}=0,6.100=60\left(g\right)\)

Câu 2:

a, Gọi CTHH cần tìm là Fe_xO_y.

Ta có: \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

Mà: m_Fe + m_O = 32 ⇒ m_O = 32 - 22,4 = 9,6 (g)

\(\Rightarrow n_O=\dfrac{9,6}{16}=0,6\left(mol\right)\)

⇒ x:y = 0,4:0,6 = 2:3

→ CTHH có dạng (Fe₂O₃)_n

\(\Rightarrow n=\dfrac{160}{56.2+16.3}=1\)

Vậy: Oxit sắt là Fe₂O₃

b, PT: \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=\dfrac{3}{2}n_{Fe}=0,6\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,6.100=60\left(g\right)\)

\(n_K=\dfrac{39}{39}=1\left(mol\right)\\ 2K+2H_2O\rightarrow2KOH+H_2\\ n_{H_2}=\dfrac{1}{2}=0,5\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{KOH}=n_K=1\left(mol\right)\\ C_{MddKOH}=\dfrac{1}{0,2}=5\left(M\right)\\ c,2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

a)

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

b)

Ta có : \(n_P = \dfrac{3,1}{31} = 0,1(mol)\)

Theo PTHH : 

\(n_{P_2O_5} = 0,5n_P = 0,05(mol)\\ n_{O_2} = \dfrac{5}{4}n_P = 0,125(mol)\)

Suy ra : 

\(m_{P_2O_5} = 0,05.142 = 7,1(gam)\\ V_{O_2} = 0,125.22,4 = 2,8(lít)\)

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

b) Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,125mol\\n_P=0,05mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{P_2O_5}=0,05\cdot142=7,1\left(g\right)\\V_{O_2}=0,125\cdot22,4=2,8\left(l\right)\end{matrix}\right.\)

1. PTHH: 4P + 5O₂  2P₂O₅

2. 

Từ PT  nO2=0,125(mol);nP2O5=0,05(mol)

đktc: VO2=0,125.22,4=2,8(l)

3. mP2O5=0,05.142=7,1(g)