\(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=>m_{H_2}=1.2=2\left(g\right)\)

\(n_{N_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)=>m_{H_2}=0,15.28=4,2\left(g\right)\)

\(n_{CO}=\dfrac{0,896}{22,4}=0,04\left(mol\right)=>m_{CO}=0,04.28=1,12\left(g\right)\)

a: \(n_{H_2}=\dfrac{V}{22.4}=1\left(mol\right)\)

b: \(n_{N_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(3.1.\left(a\right)M_P=31\left(g/mol\right);\\ M_{Fe}=56\left(g/mol\right);\\ M_{H_2}=2\left(g/mol\right);\\ M_{O_2}=32\left(g/mol\right)\\ \left(b\right).M_{P_2O_5}=31.2+16.5=142\left(g/mol\right);\\ M_{Fe_3O_4}=56.3+16.4=232\left(g/mol\right);\\ M_{HCl}=1+35,5=36,5\left(g/mol\right);\\ M_{BaO}=137+16=153\left(g/mol\right)\\ c.M_{H_2SO_4}=2+32+16.4=98\left(g/mol\right);\\ M_{ZnCl_2}=65+35,5.2=136\left(g/mol\right);\\ M_{Al_2\left(SO_4\right)_3}=27.2+\left(32+16.4\right).3=342\left(g/mol\right);\\ M_{Ca\left(OH\right)_2}=40+17.2=74\left(g/mol\right)\)

\(3.2\left(a\right).n_{CH_4}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \left(b\right).n_{CuO}=\dfrac{2}{80}=0,025\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{3,42}{342}=0,01\left(mol\right)\)

\(3.3\left(a\right).m_{HNO_3}=0,5.63=31,5\left(g\right)\\ m_{MgCl_2}=2.95=190\left(g\right)\\ \left(b\right).m_{CO_2}=\dfrac{5,6}{22,4}.44=11\left(g\right)\\ m_{H_2}=\dfrac{1,12}{22,4}.2=0,1\left(g\right)\)

\(a,\left\{{}\begin{matrix}n_C=\dfrac{4}{12}=0,25\left(mol\right)\\n_P=\dfrac{62}{31}=2\left(mol\right)\\n_{Fe}=\dfrac{42}{56}=0,75\left(mol\right)\end{matrix}\right.\\ b,\left\{{}\begin{matrix}n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mọl\right)\\n_{CO_2}=\dfrac{95,48}{44}=2,17\left(mol\right)\\n_{NaCl}=\dfrac{14,625}{58,5}=0,25\left(mol\right)\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\n_{CO}=\dfrac{12}{24}=0,5\left(mol\right)\end{matrix}\right.\)

a. \(n_C=\dfrac{4}{12}=\dfrac{1}{3}\left(mol\right)\)

\(n_P=\dfrac{62}{31}=2\left(mol\right)\)

\(n_{Fe}=\dfrac{42}{56}=0,75\left(mol\right)\)

b. \(n_{H_2O}=\dfrac{3.6}{18}=0,2\left(mol\right)\)

\(n_{CO_2}=\dfrac{95.48}{44}=2,17\left(mol\right)\)

\(n_{NaCl}=\dfrac{14.625}{58,5}=0,25\left(mol\right)\)

c. \(n_{O_2}=\dfrac{8.96}{22,4}=0,4\left(mol\right)\)

\(n_{H_2}=\dfrac{5.6}{22,4}=0,25\left(mol\right)\)

\(n_{CO}=\dfrac{12}{24}=0,5\left(mol\right)\)

a) sửa CaO

`n_(CaO)=(14/56)=0,25`mol

b)

`n_(CO_2)= 66/44=1,5` mol

c)

`n_(H_2)= (3,36)/22,4=0,15 mol`

d)

`n_(SO_3)= (22,4)/22,4=0,1 mol`

e)

`n_(Mg)=(3,55.10^23)/(6,02.10^23)=0,589mol.`

f)

tương tự e)

Ta có : \(n_{O2}=\dfrac{V}{22,4}=0,05\left(mol\right)\)

\(\Rightarrow O2=n.A=3.10^{22}\) ( phân tử )

Ta có : \(n_{SO3}=\dfrac{V}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow SO3=n.A=6.10^{22}\) ( phân tử )

Ta có : \(n_{NaOH}=\dfrac{m}{M}=0,4\left(mol\right)\)

\(\Rightarrow NaOH=n.A=2,4.10^{23}\) ( phân tử )

Ta có : \(n_{SO3}=\dfrac{m}{M}=0,405\left(mol\right)\)

\(\Rightarrow SO3=n.A=2,4381.10^{23}\) ( phân tử )

a, khối lượng của 2,5 mol CuO là:
\(m=n.M=2,5.80=200\left(g\right)\)

b, số mol của 4,48 lít khí CO2 (đktc) là:
\(n=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a.

mCuO=n.M=2,5.(1.64+1.16)= 200 mol

b.nCO₂=V/22,4=4,48:22,4=0,2 mol

\(a,n_{Fe(OH)_3}=\dfrac{5,35}{107}=0,05(mol)\\ b,n_{NH_3}=\dfrac{0,56}{22,4}=0,025(mol)\\ c,n_{CO_2}=\dfrac{11}{44}=0,25(kmol)\\ d,n_{H_2}=\dfrac{0,448}{22,4}=0,02(mol)\)