Considering the limiting reactant,how many moles of excess reactant remain after the reaction of 0.600 mol A and 1.400 mol B? A(g)+ 2 B(g)→3 C(g)

Question

Quizplus · Accepted Answer

To find out the limiting reactant we can use the mole ratio between reactants and products. For every 1 mole of A, we need 2 moles of B to produce 3 moles of C. So, to react with 0.600 mol of A we need 1.200 mol of B (0.600 mol A × (2 mol B/1 mol A) = 1.200 mol B).

Since we have 1.400 mol of B, which is greater than the amount required (1.200 mol), B is in excess and A is the limiting reactant.

The reaction consumes all of the A and produces 3/2 times the amount of C compared to A consumed. So, the amount of C produced is:

0.600 mol A × (3 mol C/1 mol A) = 1.800 mol C

The amount of B that remains after the reaction is the initial amount minus the amount used:

1.400 mol B - 1.200 mol B = 0.200 mol B

Therefore, the answer is C) 0.200 mol B, which is the amount of excess reactant remaining after the reaction.

Considering the Limiting Reactant,how Many Moles of Excess Reactant Remain