If 58 moles of NH3 are combined with 32 moles of sulfuric acid, what is the limiting reactant and how much of the excess reactant is left over?
2 NH3 + H2SO4
(NH4)2SO4
A) H2SO4, 29 mol
B) NH3, 1.0 mol
C) NH3 29 mol
D) NH3, 3.0 mol
E) H2SO4, 3.0 mol
From the balanced chemical equation it is clear that 1 mole of H2SO4 requires 2 moles of NH3. therefore, 32 moles of H2SO4 will require 64 moles of NH3. However, only 58 moles of NH3 are used in the reaction. Therefore, limiting reagent is NH3.
The excess of reactant left over = number of moles used - number of moles conusmed.
As the ideal ratio between moles of H2SO4 to NH3 is 1/2 = 0.5
The number of moles of H2SO4 consumed = 0.5 x number of moles of NH3 used
= 0.5 x 58 = 29 moles
Therefore, excess of reactant left over = 32 - 29 = 3 moles
Therefore, answer is, D) NH3, 3.0 mol
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do you find the limiting reactant? I know the formula, but I cant
remember how to round the moles to a whole number.
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> Thank you so so much this helped me huhu
Reinna Amreenah Basher Guro Mon, May 17, 2021 9:48 AM