Using 2H2+ Oz--> 2H20. Determine the limiting reactant when 19g of hydrogen reacts with 19g of oxygen

Question

Well I don’t understand what this means or what the answer is.. I just need to figure everything out before tomorrow

Answer 1

The balanced equation is very important as it shows you the mole relationships between reactants and products. In this question, you're given masses so we can't dive into the equation right away; we must convert the masses given to moles.

The only way to convert the mass of a substance to moles, is to use the molar mass. The molar mass is calculated by using the masses indicated on the periodic table.

Let's start by converting the masses given to moles. The mass of H₂ given is 19 g. TO calculate the molar mass of H₂ = 2 x H = 2x 1.008 g = 2.016 g. This means that 1 mole of H₂ = 2.016 g

We'll do the same for O₂.......molar mass = 2 x O = 2 x 15.9994 g = 31.9988 g. This means that 1 mole of O₂ = 31.9988 g

To convert from mass to moles, we divide by the molar mass of each substance.

For H₂ moles = 19 g x 1 mol/2.016 g = 9.4 mols H₂

For O_2, moles = 19 g x 1 mol/31.9988 g = 0.59 mols O₂

This means that for the reaction, we have 9.4 mols H₂ and 0.59 mols O₂. Since we have moles, we can now use the balanced equation. The balanced equation says:

2 H₂ + O₂ → 2 H₂O

This means that for the reaction, we need twice as many moles of H₂ than O₂ because the ration between them shows 2 mol H₂: 1 mol O₂.

We have 0.59 mols O₂ and 9.4 mols H₂, so we have more than enough H₂ for the reaction, since 2 x 0.59 = 1.18 and we have 9.4 mols.

We could also look at it from hydrogen's angle and say that we need half as much oxygen. We have 9.4 mols of hydrogen, and 9.4 mols/2 = 4.7 mols. We only have 0.59 mols O₂ so it's not enough to react with all the hydrogen that's there. It will be used up first.

This means that O₂ is the limiting reactant since it will be used up completely leaving some amount of H₂ behind (excess).