Answer:
The number of people in that specific area.
Explanation:
<u>Answer:</u> The mass of iron (II) oxide that must be used in the reaction is 30.37
<u>Explanation:</u>
The given chemical reaction follows:

By Stoichiometry of the reaction:
When 635 kJ of energy is released, 6 moles of iron (II) oxide is reacted.
So, when 44.7 kJ of energy is released,
of iron (II) oxide is reacted.
Now, calculating the mass of iron (II) oxide by using the equation:

Moles of iron (II) oxide = 0.423 moles
Molar mass of iron (II) oxide = 71.8 g/mol
Putting values in above equation, we get:

Hence, the mass of iron (II) oxide that must be used in the reaction is 30.37
3H2+N2→2NH3
here h2 H2 is limiting reagent
6moles of H2 will give 4moles of NH3
V of NH3 produced=22.4*4L
Answer:
1) 4Fe + 3O2 → 2Fe2O3
2) H2 + Cl2 → 2HCl
3) 2Ag + H2S → Ag2S + H2
4) CH4 + 2O2 → CO2 + 2H2O
5) 2HgO → 2Hg + O2
6) 2Co + 3H2O → Co2O3 + 3H2