Given a reaction between 3 moles of A and 4 moles of B to produce C and D, if you have 5 moles of A and 6 moles of B initially, which reactant is the limiting reactant, and how many moles of each product will be formed?
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To determine the limiting reactant, we need to compare the moles of A and B with the stoichiometric ratio of the reaction.
The reaction is:
3A + 4B -> C + D
From the stoichiometry, we can see that for every 3 moles of A, we need 4 moles of B.
Initially, we have 5 moles of A and 6 moles of B.
To see how many moles of C and D will be formed, we need to determine which reactant will be completely consumed.
Let's check the ratio of moles of A to B based on the stoichiometry:
Moles of A / Stoichiometric ratio of A = 5 / 3 = 1.67
Moles of B / Stoichiometric ratio of B = 6 / 4 = 1.5
Comparing the two ratios, we can see that B is the limiting reactant because it has the lower moles-to-stoichiometric ratio.
Therefore, B is the limiting reactant in this reaction.
To find the number of moles of C and D produced, we need to use the stoichiometry:
For every 4 moles of B, we produce 1 mole of C and 1 mole of D.
Since B is the limiting reactant, we can determine the number of moles of C and D based on the number of moles of B.
Moles of C = (Moles of B) * (Number of moles of C / Number of moles of B) = 6 * (1/4) = 1.5 moles
Moles of D = (Moles of B) * (Number of moles of D / Number of moles of B) = 6 * (1/4) = 1.5 moles
Therefore, 1.5 moles of C and 1.5 moles of D will be formed when 5 moles of A and 6 moles of B react.
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