A student has 10.0mL of phosphate solution that contains 1.0M PO4{3-}.The Student reacts this with excess ammonium & magnesium under basic conditions to produce 2.2034g of struvite. Calculate the % yield.
Mg^2+ + NH3 + 7H2O + PO4^3- ==> MgNH4PO4.6H2O + OH^-
mols PO4^3- = M x L = 1.0 x 0.010 = 0.010
mols MgNH4PO4.6H2O = 0.010
g MgNH4PO4.6H2O = mols x molar mass = 0.010 x 245.41 = ? This is the theoretical yield.
%yield = (2.2034/?)*100 = ???
yes that's the right answer
To calculate the percent yield, we need to compare the actual yield to the theoretical yield.
First, we need to determine the limiting reagent, which is the reactant that will be completely consumed and determines the maximum amount of product that can be formed. In this case, the reactant that will limit the formation of struvite is either ammonium or magnesium.
Since the problem states that there is an excess of both ammonium and magnesium, we can assume that the limiting reagent is the phosphate solution containing PO4{3-}.
To calculate the moles of PO4{3-} in the solution, we can use the formula:
moles = concentration (molarity) × volume (in liters)
Given:
Concentration (molarity) of PO4{3-} = 1.0 M
Volume of phosphate solution = 10.0 mL (which can be converted to liters by dividing by 1000)
moles of PO4{3-} = 1.0 M × 10.0 mL ÷ 1000 = 0.01 moles
Now, let's calculate the theoretical yield of struvite using stoichiometry. The balanced equation for the reaction is:
(NH4)2PO4 + 6Mg(OH)2 + 3H2O → MgNH4PO4·6H2O
From the balanced equation, we can see that the ratio between (NH4)2PO4 and MgNH4PO4·6H2O is 1:1. The molar mass of MgNH4PO4·6H2O is:
MgNH4PO4·6H2O = 2 × (1 + 14) + 31 + 16 × 4 + 6 × 2 + 16 × 6 = 245 g/mol
The theoretical yield can be calculated using the formula:
theoretical yield = moles of limiting reagent × molar mass of product
theoretical yield = 0.01 moles × 245 g/mol = 2.45 g
Finally, we can calculate the percent yield using the formula:
percent yield = (actual yield / theoretical yield) × 100
percent yield = (2.2034 g / 2.45 g) × 100 = 89.9%
Therefore, the percent yield of this reaction is 89.9%.