What is the limiting reagent for 6HCl + Fe2O3 = 2FeCl3 + 3H2O ? 6HCl has 18.44 moles and Fe2O3 has 2.91 moles. Also how many moles would FeCl3 produce?
To determine the limiting reagent and calculate the moles of FeCl3 produced, we need to compare the moles of reactants to their stoichiometric coefficients.
Given:
- 6HCl: 18.44 moles
- Fe2O3: 2.91 moles
Let's start by calculating the moles of FeCl3 that could be formed from each reactant:
From the balanced equation:
6HCl + Fe2O3 → 2FeCl3 + 3H2O
According to the stoichiometry, for every 6 moles of HCl, 2 moles of FeCl3 are produced. Thus, we can use the ratio of 6 moles HCl to 2 moles FeCl3 to determine the moles of FeCl3 produced from 18.44 moles of HCl:
(18.44 moles HCl) x (2 moles FeCl3 / 6 moles HCl) = 6.1467 moles FeCl3
Similarly, for every 1 mole of Fe2O3, 2 moles of FeCl3 are produced. We can use this ratio to determine the moles of FeCl3 produced from 2.91 moles of Fe2O3:
(2.91 moles Fe2O3) x (2 moles FeCl3 / 1 mole Fe2O3) = 5.82 moles FeCl3
Now, let's compare the moles of FeCl3 produced from each reactant:
Moles of FeCl3 from HCl: 6.1467 moles
Moles of FeCl3 from Fe2O3: 5.82 moles
Since the moles of FeCl3 produced from Fe2O3 (5.82 moles) are lower than those from HCl (6.1467 moles), Fe2O3 is the limiting reagent. This means that Fe2O3 will be completely consumed, and the amount of product formed will be limited by the amount of Fe2O3 available.
Therefore, the limiting reagent is Fe2O3, and it would produce 5.82 moles of FeCl3.