Question

you must calculate how many moles of product each reactant can form (assuming there is enough of all other reactants); the limiting reactant is the reactant that when used up produces the least amount of product (in moles); and in this method, you can use any product of your choice. 2 h₂ (g) + 1 o₂ (g) → 2 h₂o (g) 3 moles of o₂ 5 moles of h₂ if you used five moles of h₂, how many moles of h₂o (g) would form (assuming you have enough o₂)? if you used three moles of o₂, how many moles of h₂o (g) would form (assuming you have enough o₂)? what is the limiting reactant? o₂ h₂ h₂o

Explanation:

Step1: Calculate moles of \(H_2O\) from \(H_2\)

From the balanced equation \(2H_2(g)+O_2(g)\to2H_2O(g)\), the mole ratio of \(H_2\) to \(H_2O\) is \(2:2 = 1:1\).
If \(n(H_2)=5\space mol\), then \(n(H_2O)=\frac{2\space mol\space H_2O}{2\space mol\space H_2}\times5\space mol\space H_2 = 5\space mol\space H_2O\)

Step2: Calculate moles of \(H_2O\) from \(O_2\)

The mole ratio of \(O_2\) to \(H_2O\) is \(1:2\).
If \(n(O_2) = 3\space mol\), then \(n(H_2O)=\frac{2\space mol\space H_2O}{1\space mol\space O_2}\times3\space mol\space O_2=6\space mol\space H_2O\)

Step3: Determine the limiting reactant

Since \(H_2\) produces fewer moles of \(H_2O\) (5 mol vs 6 mol from \(O_2\)), \(H_2\) is the limiting reactant.

Answer:

When using 5 moles of \(H_2\) (assuming enough \(O_2\)): \(5\space mol\) of \(H_2O\) forms.
When using 3 moles of \(O_2\) (assuming enough \(H_2\)): \(6\space mol\) of \(H_2O\) forms.
Limiting reactant: \(H_2\)