The first step of the synthesis is described by the reaction below. When 1.750 g of Fe(NH4)2(SO4)2 6H2O is mixed with 13 mL of 1.0 M H2C2O4, the theoretical yield of FeC2O42H2O is____
grams.
Fe(NH4)2(SO4)26H2O(s) + H2C2O4(aq)---->
FeC2O42H2O(s) + (NH4)2SO4(aq) + H2SO4(aq) + 4 H2O(l)
you have grams of product and you have moles from the equation so you can find the mass of product theoretically formed.
usually you would find the limiting reagent but here it's a 1:1 ratio so it doesn't matter.
1.750g Fe(NH4)2(SO4)2 6H2O x(1 molFe(NH4)2(SO4)2 6H2O/__g)(1mol FeC2O42H2O/1mol Fe(NH4)2(SO4)2 6H2O )= _________moles of product formed from Fe(NH4)2(SO4)2 6H2O
13 mL(1L/1000ml)(1mol H2C2O4/L)( 1mol FeC2O42H2O/1mol H2C2O4)= _______moles of product formed from H2C2O4
_____________
The reactant that produces the least moles of product is the limiting reagent.
Take that number of moles of product formed and use it to form the theoretical yield of product.
(____moles of FeC2O42H2O)(____g FeC2O42H2O/ 1 mol)= _____________g FeC2O42H2O produced.
If you need clarification on anything just ask.
To find the theoretical yield of FeC2O42H2O, you need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
First, convert the mass of Fe(NH4)2(SO4)2 6H2O to moles using its molar mass. The molar mass of Fe(NH4)2(SO4)2 6H2O is:
1 mol Fe(NH4)2(SO4)2 6H2O = 392.14 g
Moles of Fe(NH4)2(SO4)2 6H2O = 1.750 g / 392.14 g/mol
Next, calculate the moles of H2C2O4 using its molarity and volume:
1 M = 1 mol/L
Moles of H2C2O4 = 1.0 mol/L * (13 mL / 1000 mL)
Now, using the balanced equation, determine the mole ratio between Fe(NH4)2(SO4)2 6H2O and FeC2O42H2O. From the equation, it is clear that 1 mole of Fe(NH4)2(SO4)2 6H2O reacts to form 1 mole of FeC2O42H2O.
Since the mole ratio is 1:1, the number of moles of FeC2O42H2O formed is equal to the number of moles of Fe(NH4)2(SO4)2 6H2O.
Finally, convert the moles of FeC2O42H2O to grams using its molar mass. The molar mass of FeC2O42H2O is 548.21 g/mol.
Theoretical yield of FeC2O42H2O = Moles of FeC2O42H2O * Molar mass of FeC2O42H2O
Remember to round the final answer to the appropriate number of significant figures.