Здравствуйте.
<span>Р</span>₂<span>О</span>₅ <span>+ 3Н</span>₂<span>О = 2Н</span>₃<span>РО</span>₄<span>
</span><span>CaCO</span>₃<span> + 2HCl = H</span>₂<span>O + CO</span>₂<span> + CaCl</span>₂<span>
</span>
BaCl₂ + H₂SO₄ <span>= BaSO</span>₄ <span>+ 2HCl
</span>
Fe + CuSO₄(разб.) = FeSO₄ + Cu↓
<span>Cu + 2AgNO</span><span>₃ </span><span>= Cu(NO</span><span>₃</span><span>)</span><span>₂</span><span> + 2Ag
</span>
<span>Zn + Pb(NO</span><span>₃</span><span>)</span><span>₂ </span><span>= Pb + Zn(NO</span><span>₃</span><span>)</span><span>₂
</span>
<span>3КСNS + FеСl</span><span>₃</span><span> = Fe(CNS)</span><span>₃</span><span> + </span><span>3КСl
</span>
BaCl₂ + K₂Cr₂O₇ = 2KCl + BaCr₂O₇*
<span>NiSO</span><span>₄</span><span> + 2NaOH = Ni(OH)</span><span>₂</span><span> + Na</span><span>₂</span><span>SO</span><span>₄
</span>
<span>2NaOH + CuSO</span><span>₄ </span><span>= Na</span><span>₂</span><span>SO</span><span>₄ </span><span>+ Cu(OH)</span><span>₂</span>
2KCLO3=2KCL+3O2
1)n(KCLO3);n=m/M;n(KCLO3)=100/32=3,125моль
Подставляем 3,125 в уравнение над бертолетовой солью
2)n(O2)=3,125/2=X/3 X=4,6875 моль
3)V(O2);V=n*Vm;V(O2)=4,6875*22,4=105 л
Ответ:V(O2)=105 л
W=n*Ar/Mr*100%
Mr= 39+1+12+16*3=100
w(K)=1*39/100 *100%= 39%
w(H)=1/100 *100%=1%
w(C)=12/100 *100%=12%
w(O)= 16*3/100 *100%=48%
При горении сложных веществ образуются оксидЫ! например:
CH4+2O2=CO2+2H2O