Asked by dody
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?
Answers
Answered by
DrBob222
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.
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.
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