Wen 4Ag + 2H2S + 02 -> 2Ag2S + 2H20 and the reactants inclde 12 mol Ag, 6 mole of H2S and 96 G of 02 the limiting reactant is?
To determine the limiting reactant, we need to compare the amount of each reactant to their stoichiometric coefficients in the balanced chemical equation.
Let's start by finding the moles of each reactant:
1. Silver (Ag):
Given 12 moles of Ag, we have:
12 moles Ag
2. Hydrogen sulfide (H2S):
Given 6 moles of H2S, we have:
6 moles H2S
3. Oxygen (O2):
Given 96 g of O2, we need to convert it to moles. The molar mass of O2 is 32 g/mol, so:
96 g O2 * (1 mol O2 / 32 g O2) = 3 moles O2
Now, let's compare the moles of each reactant to their stoichiometric coefficients in the balanced equation.
From the balanced equation:
4 Ag + 2 H2S + O2 -> 2 Ag2S + 2 H2O
Comparing the coefficients:
Ag (Silver):
For 4 moles of Ag, we need 1 mole of O2.
H2S (Hydrogen sulfide):
For 2 moles of H2S, we need 1 mole of O2.
O2 (Oxygen):
To completely react, 1 mole of O2 is required.
Based on the stoichiometric ratios, we can see that both Ag and H2S require 1 mole of O2 for reaction. However, we only have 3 moles of O2 available. This means we have a limited amount of O2, making it the limiting reactant.
Thus, the limiting reactant is O2.
Convert 96 g O2 to moles. moles = grams/molar mass = 96/16 = 6
Now, using the coefficients in the balanced equation, convert each of the reactants to any one of the products. We may as well use Ag2S.
12 moles Ag x (2 moles Ag2S/4 moles Ag) = 12*2/4 = 6 moles Ag2S.
6 moles H2S x (2 mols Ag2S/2 moles H2S) = 6*2/2 = 6 moles Ag2S.
6 moles O2 x (2 moles Ag2S/1 mole O2) = 6*2/1 = 12.
These numbers for moles Ag2S formed don't agree. The correct value is ALWAYS the smallest; in this case, it is Ag and/or H2S.