Respuesta :
57.5 % is the percent yield if 10.9 g of H₃PO₄ is formed.
Explanation:
Balanced equation for the reaction:
Ca₃(PO₄)₂ (s) + 2H₂SO₄ (aq) → H₃PO₄ (aq) + 2CaSO₄ (aq)
data given:
mass of Ca₃(PO₄)₂ = 19.9 grams
mass of H₂SO₄, = 54.3 grams
mass of H₃PO₄ produced = 10.9 grams (actual yield)
percent yield=?
atomic mass of Ca₃(PO₄)₂ = 310.17 grams/mole
atomic mass of H₂SO₄ = 98.07 grams/mole
number of moles is calculated as:
number of moles = [tex]\frac{mass}{atomic mass of one mole}[/tex]
putting the values in the above equation:
for Ca₃(PO₄)₂ = [tex]\frac{19.9}{310.17}[/tex]
= 0.064 moles
moles of H₂SO₄ = [tex]\frac{54.3}{98.07}[/tex]
= 0.55 moles
from the reaction, it can be found that the limiting reagent is Ca₃(PO₄)₂
1 mole of Ca₃(PO₄)₂ reacts to form 1 mole of phosphoric acid
0.064 moles of Ca₃(PO₄)₂ will produce x moles of phosphoric acid
0.064 moles of phosphoric acid produced
mass = number of moles x atomic mass of H3PO4
=0.064 x 97.994
= 6.27 grams (theoretical yield)
FORMULA FOR PERCENT YIELD:
percent yield = [tex]\frac{actual yield}{theoretical yield}[/tex] x 100
= [tex]\frac{6.27}{10.9}[/tex] x 100
= 57.5 %
