Bài 1 :

a) \(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\)

b)

\(n_{O_2} = \dfrac{1}{2}n_{H_2} = \dfrac{1}{2}. \dfrac{6,72}{22,4} = 0,15(mol)\\ \Rightarrow V_{O_2} = 0,15.22,4 = 3,36(lít)\\ \Rightarrow m_{O_2} = 0,15.32 = 4,8(gam)\)

Bài 2 :

\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\)

Vì  :\( \dfrac{n_{H_2}}{2} = \dfrac{8,4}{22,4.2} = 0,1875 > \dfrac{n_{O_2}}{1} = \dfrac{2,8}{22,4} =0,125\) nên Hidro dư

\(n_{H_2O} = 2n_{O_2} = 0,125.2 = 0,25(mol)\\ \Rightarrow m_{H_2O} = 0,25.18 = 4,5(gam)\)

\(a) 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ b)V_{O_2} = \dfrac{1}{2}V_{H_2} = 0,7(lít)\\ V_{không\ khí} = \dfrac{V_{O_2}}{20\%} = \dfrac{0,7}{20\%} = 3,5(lít)\\ c) n_{H_2O} = n_{H_2} = \dfrac{1,4}{22,4} = 0,0625(mol)\\ m_{H_2O} = 0,0625.18 = 1,125(gam)\)

2H2+O2-to>2H2O

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

n H2=0,1 mol

=>m H2OI=0,1.18=1,8g

=>Vkk=0,05.22,4.5=5,6l

câu này là câu a hay b z bn ?

- Bạn ơi, 5,6 lít của nước hay hiđro

Hidro nha b.Từ nước thay = khí.Viết nhầm á

2H2+O2-to>2H2O

0,2----0,1-----0,2

n H2=0,2 mol

=>m H2O=0,2.18=3,6g

=>Vkk=0,1.22,4.5=11.2l

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

PTHH: 2H₂ + O₂ ----t^o----> 2H₂O

Mol:      0,2    0,1                   0,2

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

b, \(V_{O_2}=0,1.22,4=2,24\left(l\right)\Rightarrow V_{kk}=2,24.5=11,2\left(l\right)\)

a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

b, \(n_{CH_4}=\dfrac{28}{22,4}=1,25\left(mol\right)\)

\(n_{CO_2}=n_{CH_4}=1,25\left(mol\right)\Rightarrow m_{CO_2}=1,25.44=55\left(g\right)\)

c, \(n_{O_2}=2n_{CH_4}=2,5\left(mol\right)\Rightarrow V_{O_2}=2,5.22,4=56\left(l\right)\)

Zn+2HCl->Zncl2+H2

0,4----0,8----0,4----0,4

n Zn=0,4 mol

VH2=0,4.22,4=8,96l

m ZnCl2=0,4.136=54,4g

2H2+O2-to>2H2O

0,4------0,2----0,4

n O2=0,2 mol

=>pứ hết 

=>m H2O=0,4.18=7,2g

a.b.\(n_{Zn}=\dfrac{26}{65}=0,4mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,4                      0,4       0,4       ( mol )

\(m_{ZnCl_2}=0,4.136=54,4g\)

\(V_{H_2}=0,4.22,4=8,96l\)

c.\(n_{O_2}=\dfrac{4,48}{22,4}=0,2mol\)

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

0,4   = 0,2                       ( mol )

0,4        0,2            0,4        ( mol )

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

2H2+O2-to>2H2O

0,25---0,125-----0,25

n H2=0,25 mol

=>m H2O=0,25.18=4,5g

=>Vkk=0,125.22,4.5=14l

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\\ Mol:0,25\rightarrow0,125\rightarrow0,25\\ V_{kk}=0,125.5.22,4=14\left(l\right)\\ m_{H_2O}=0,25.18=4,5\left(g\right)\)

a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

b, Ta có: \(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,2.232=46,4\left(g\right)\)

c, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,4\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\) \(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{21\%}\approx42,67\left(l\right)\)

d, PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

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

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,4}{4}\), ta được Fe₃O₄ dư.

Theo PT: \(n_{Fe_3O_4\left(pư\right)}=\dfrac{1}{4}n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow n_{Fe_3O_4\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

a. \(3Fe+2O_2\rightarrow Fe_3O_4\)

b. Số mol Fe: \(n=\dfrac{m}{M}=\dfrac{33,6}{56}=0,6\left(mol\right)\)

          PTHH: \(3Fe+2O_2\rightarrow Fe_3O_4\)

 Theo PTHH:  \(3\)        \(2\)            \(1\)           (mol)

       Theo đề:  \(0,6\)           \(\rightarrow0,2\)          (mol)

Kl của \(Fe_3O_4\) là: \(m=n\cdot M=0,2\cdot\left(56\cdot3+16\cdot4\right)=736\left(g\right)\)