The question is incomplete. The complete question is
Two solutions are combined in a beaker. One solution contains 500.0 g of potassium phosphate and the other contains 500.0 g of calcium nitrate. A double displacement reaction occurs. What mass of each of the following substances is present when the reaction stops. A) potassium phosphate remaining B) calcium nitrate g remaining C) calcium phosphate formed D) potassium nitrate g formed
Answer:
a)84.91g
b)8.20g
c)316.4g
d)616.73g
Step-by-step explanation:
The equation of the reaction:
2K3PO4(aq) + 3Ca(NO3)2 (aq)-------> 6KNO3(aq) + Ca3(PO4)2(s)
Molar mass of potassium phosphate= 212.27 g/mol
Amount of potassium phosphate= 500/212.27= 2.4 moles
Molar mass of calcium nitrate= 164.088 g/mol
Amount of calcium nitrate= 500/164.088=3.05moles
a) amount of potassium phosphate reacted according to reaction equation= 2 moles
Amount of potassium phosphate remaining= 2.4-2=0.4 moles
Mass of potassium phosphate remaining= 0.4×212.27=84.91g
b) Amount of calcium nitrate reacted according to reaction equation=3
Amount of calcium nitrate remaining=3.05-3= 0.05
Mass of calcium nitrate remaining= 0.05×164.088= 8.20g
c) since calcium nitrate is the limiting reactant, we use to estimate the mass of products formed.
From the reaction equation,
3 moles of calcium nitrate yields 1 mole of calcium phosphate
3.05 moles of calcium nitrate yields 3.05/3 = 1.02 moles of calcium phosphate
Molar Mass of calcium phosphate= 310.18 g/mol
Mass of calcium phosphate produced= 1.02×310.18= 316.4g
d)
3 moles of calcium nitrate yields 6 moles of potassium nitrate
3.05 moles of calcium nitrate yields 3.05×6/3= 6.1 moles of potassium nitrate
Molar mass of potassium nitrate = 101.1032 g/mol
Mass of potassium nitrate formed= 6.1× 101.1032= 616.73g