Verified answer

(1) Moles of O2(g) needed to completely react with 2.00 moles of NH3(g).

-Use a conversion factor fraction.

-Start with moles of NH3, then the molar ratio of O2 to NH3.

[2 mol NH3/1] x [5 mol O2 / 4 mol NH3] = 2.5 moles of O2

(as you can see, the variable of mol NH3 cancel out leaving mol O2)

(2) Moles of H2O formed.

-Use the same process as before.

[2 mol NH3 /1] x [6 mol H2O / 4 mol NH3] = 3 moles of H2O

(3) grames of NO(g) formed with 2 moles of NH3.

-Again use a conversion factor fraction.

[2 mol NH3 /1] x [4 mol NO / 4 mol NH3] x [30.01 g NO / 1 mol NO] = 60.01 grams of NO

*the balanced equation is giving the ratio between the different chemicals

*in using the conversion factor fraction, start off with what you have, and look to where you what to go

hope this helps!

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Four moles of ammonia combine with 5 moles of oxygen gas to make four moles of nitrogen monoxide and six moles of water vapor.

The relative numbers of moles of reactants and products are given by the coefficients in the balanced chemical equation. The coefficients will tell you any ratio of molecules of reactant and product.

If 2.0 moles of NH3 react with excess O2, then 2 moles of NO will be produced, since they are in a 1:1 mole ratio as indicated by the balanced chemical equation. Then simply run the number of moles through a conversion factor to find the mass.

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