С3H8 + 5O2 = 3CO2 + 4H2O
m(C3H8)=7
M(C3H8)=36+8=44
Mol(C3H8)=7/44=0.16 где то
Пропорция
0.16/1=x/3
x=0.48
M(CO2)=44
m(CO2)=44*0.48=21.12
V(CO2)=22.4*0.48=10.752
N(CO2)=0.48*6.02=2.8896 * 10^23
1. Na + O2 → Na2O
2. 2 Li + 2 H2O → 2 LiH + H2
3. Mg + HNO3 → Mg(NO3)2 + N2O + H2O (ето не точно)
4.
5.
6. 2 Al + 6 NaOH → 3 H2 + 2 Na3AlO3
7.
8.Cu + H2O → CuO + H2
Дано:
m (H2SO4)=39.2 г
_______
m (Ba(NO3)2)-?
M (H2SO4)=98 г / моль
М (Ba(NO3)2)=261 г / моль
Решение :
Ba (NO3)2 + H2SO4 -> BaSO4 + 2HNO3
1) n (H2SO4)= m / M = 39.2г / 98 г / моль = 0.4 моль
2) х/1 моль = 0.4 / 1 моль
х = 0.4 × 1 / 1 = 0.4 моль ( n(Ba(NO3)2))
3)m в-ва(Ba(NO3)2 = n × M = 0.4 × 261 г / моль = 104.4 г
Ответ:m (Ba(NO3)2)= 104.4г
А) Mg + 2HBr = <span>H2</span> + <span>MgBr2</span>BAO<span> + </span>2<span>HNO3</span><span> = </span><span>BA(NO3)2</span><span> + </span><span>H2O
</span>2NaOH<span> + </span><span>H2S</span><span> = </span><span>Na2S</span><span> + </span>2<span>H2O</span>
б) 2<span>Li2O</span> + 7<span>H2SO4</span> = 4<span>LiO4</span> + 7<span>SO2</span> + 7<span>H<span>2
</span></span>Fe(OH)2<span>+ </span>2<span>HNO3</span><span> = </span><span>Fe(NO3)2</span><span> + </span>2<span>H2O
</span>3<span>K2SiO3</span><span> + </span>2<span>H3PO4</span><span> = </span>2<span>K3PO4</span><span> + </span>3<span>H2SiO<span>3</span></span>