Also 0,3 moles nitric oxide.
2Al + 3O2 --> 2Al2O3 0.78 moles O2 (2 moles Al2O3/3 moles O2) 0.52 moles Al2O3 produced ===========================( assuming oxygen limits )
2Cl2 + 7O2 --> 2Cl2O7So for every 2 moles of dichlorine heptoxide formed, 7 moles of oxygen are reacted.69.1 moles Cl2O7 * (7 moles O2 / 2 moles Cl2O7) = 241.85 moles O2
Using the molar mass of nh3, we find that we have 2.5 moles of nh3. Since 3 moles of h2o are produced per 2 moles of nh3, we see that we will produce 3.75 moles of h2o. This is equivalent to around 3.79 g.
When methane undergoes complete combustion, the equation for the reaction is CH4 + 2 O2 -> CO2 + 2 H2O. This shows that the number of moles of carbon dioxide formed are the same as the number of moles of methane reacted, so that 14 moles of carbon dioxide will be formed from 14 moles of methane.
150.0 g O2 x 1 mole O2/32 g O2 = 4.688 moles O2
2Al + 3O2 --> 2Al2O3 0.78 moles O2 (2 moles Al2O3/3 moles O2) 0.52 moles Al2O3 produced ===========================( assuming oxygen limits )
1.5 moles of O2 react with 2 moles of Al 2Al + 1.5 O2 = Al2O3.
49.7g O2
2Cl2 + 7O2 --> 2Cl2O7So for every 2 moles of dichlorine heptoxide formed, 7 moles of oxygen are reacted.69.1 moles Cl2O7 * (7 moles O2 / 2 moles Cl2O7) = 241.85 moles O2
Using the molar mass of nh3, we find that we have 2.5 moles of nh3. Since 3 moles of h2o are produced per 2 moles of nh3, we see that we will produce 3.75 moles of h2o. This is equivalent to around 3.79 g.
The answer is 2 moles.
The most straightforward reaction for the formation of SO3 from SO2 is 2 SO2 + O2 => 2 SO3. If this is the actual reaction for the formation, 3 moles of SO3 are formed from 3 moles of SO2.
When methane undergoes complete combustion, the equation for the reaction is CH4 + 2 O2 -> CO2 + 2 H2O. This shows that the number of moles of carbon dioxide formed are the same as the number of moles of methane reacted, so that 14 moles of carbon dioxide will be formed from 14 moles of methane.
150.0 g O2 x 1 mole O2/32 g O2 = 4.688 moles O2
Only when 5.5 mole O2 react with 11 mole H2, then 11 mole H2O are formed.
2H2 + O2 --> 2H2OAs you can see by the balanced reaction, for every 1 mole of oxygen used, 2 moles of water are formed. Also notice that for every 1 mole of oxygen used, you need 2 moles of hydrogen to produce the 2 moles of water. So in your case 110 moles of oxygen would produce 220 moles of water & would also require 220 moles of hydrogen (which you have in excess since you have 230 moles of hydrogen). So 220 moles of water are the most that can be formed.
Balanced equation. 2S + 3O2 --> 2SO3 1.32 moles O2 (2 moles SO3/3 moles O2) = 0.880 moles sulfur trioxide produced ========================