Đáp án:

Giải thích các bước giải:

$1/$

$PTPƯ:2H_2+O_2\buildrel{{t^o}}\over\longrightarrow$ $2H_2O$

$a,n_{H_2}=\frac{5,6}{22,4}=0,25mol.$

$Theo$ $pt:$ $n_{O_2}=\frac{1}{2}n_{H_2}=0,125mol.$

$⇒V_{O_2}=0,125.22,4=2,8l.$

$⇒m_{O_2}=0,125.32=4g.$

$b,Theo$ $pt:$ $n_{H_2O}=n_{H_2}=0,25mol.$

$⇒m_{H_2O}=0,25.18=4,5g.$

$2/$

$PTPƯ:Fe_2O_3+3H_2\buildrel{{t^o}}\over\longrightarrow$ $2Fe+3H_2O$

$n_{Fe_2O_3}=\frac{3,2}{160}=0,02mol.$

$Theo$ $pt:$ $n_{Fe}=2n_{Fe_2O_3}=0,04mol.$

$⇒m_{Fe}=0,04.56=2,24g.$

$b,Theo$ $pt:$ $n_{H_2}=3n_{Fe_2O_3}=0,06mol.$

$⇒V_{H_2}=0,06.22,4=1,344l.$

$3/$

$a,PbO+H_2\buildrel{{t^o}}\over\longrightarrow$ $Pb+H_2O$

$b,CuO+H_2\buildrel{{t^o}}\over\longrightarrow$ $Cu+H_2O$

$c,Fe_2O_3+3H_2\buildrel{{t^o}}\over\longrightarrow$ $2Fe+3H_2O$

$d,ZnO+H_2\buildrel{{t^o}}\over\longrightarrow$ $Zn+H_2O$

$e,Fe_3O_4+4H_2\buildrel{{t^o}}\over\longrightarrow$ $3Fe+4H_2O$

$f,MgO+H_2\buildrel{{t^o}}\over\longrightarrow$ $Mg+H_2O$

chúc bạn học tốt!

1/ nH2=5,6/22,4=0,25mol

2H2 + O2 → 2H2O

0,25--->0.125--->0,25 (mol)

a/ VO2=0,125x22,4=2,8 lít

mO2= 0.125x16=2g

b/mH2O=0.25x18=2,25g

2/

 nFe2O3=3,2/160=0,02mol

Fe₂O₃+3H₂→2Fe+3H₂O

0,02 -->0,06-->0,04(mol)

a/mFe=0,04x56=2.24g

b/VH2=0,06x22,4=1,344 lít

3/

a/H₂+PbO⟶H₂O+Pb (ĐK: nhiệt độ)

b/CuO + H2 → Cu + H2O(ĐK: nhiệt độ)

c/Fe₂O₃+3H₂→2Fe+3H₂O(ĐK: nhiệt độ)

d/H₂+ZnO⟶H₂O+Zn(ĐK: ko có )

e/4H₂+Fe₃O₄→3Fe+4H₂O(ĐK: nhiệt độ)

f/H₂O+Mg→H₂+MgO(ĐK: nhiệt độ)