Question

Suppose that, for reaction CuO (s) + H2SO4 (aq) → CuSO4 (aq) + H2O (l) you could not find the bottle of 7 M H2SO4 so you added 5.00 mL of the 1.00 M H2SO4 instead. How would this impact your final yield of Copper. (Show with calculations how this would impact the limiting reagent.

Answer 1

CuO (s) + H2SO4 (aq) → CuSO4 (aq) + H2O (l)

number of moles of H2SO4 = 5*1 = 5 mmol/L

from the equation,

1 mole of H2SO4 produces 1 mole of CuSO4

5 mmol of H2SO4 produces 5 mmol of CuSO4

So 5 mmol is exact yield

similarly,

number of moles of H2SO4 originally assume = 7 * 5 = 35 mmol/L

From the equation,

1 mole of H2SO4 produces 1 mole of CuSO4

35 mmol of H2SO4 produces 35 mmol of CuSO4

35 mmol is the assumed yield

percent of yield = 5mmol/35mmol*100 = 14.29%

so we will get 14.29% of CuSO4 if we add 1M H2SO4 insted of 7M H2SO4.