If 0.273 moles of ferrous ammonium sulfate (Fe(NH4)2(SO4)2 . 6H2O) and an excess of all other reagents are used in a synthesis of K2[Fe(C2O4)3] . 3H2O, how many grams of product will be obtained if the reaction gives a 100% yield?
Please help i found the molar mass of (Fe(NH4)2(SO4)2 . 6H2O) = 491.1g/mol and
K2[Fe(C2O4)3] . 3H2O = 392.1 g/ mol.
This is a stoichiometry problem. Here is a site that show you how to solve all "simple" stoichiometry problems (but not limiting reagent problems).
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the mass of the product obtained, we need to calculate the moles of the product formed and then multiply it by the molar mass.
1. Start with the given amount of ferrous ammonium sulfate (Fe(NH4)2(SO4)2 . 6H2O):
Moles of Fe(NH4)2(SO4)2 . 6H2O = 0.273 moles
2. The balanced equation for converting Fe(NH4)2(SO4)2 . 6H2O to K2[Fe(C2O4)3] . 3H2O is:
3 Fe(NH4)2(SO4)2 . 6H2O + 10 K2C2O4 + 24 H2SO4 -> 6 K2[Fe(C2O4)3] . 3H2O + 4 (NH4)2SO4 + 24 H2O
3. From the balanced equation, we can see that the stoichiometric ratio between Fe(NH4)2(SO4)2 . 6H2O and K2[Fe(C2O4)3] . 3H2O is 3:6 (reduced to 1:2).
4. Convert moles of Fe(NH4)2(SO4)2 . 6H2O to moles of K2[Fe(C2O4)3] . 3H2O:
Moles of K2[Fe(C2O4)3] . 3H2O = (0.273 moles Fe(NH4)2(SO4)2 . 6H2O) * (2/3) = 0.182 moles
5. Finally, calculate the mass of K2[Fe(C2O4)3] . 3H2O:
Mass of K2[Fe(C2O4)3] . 3H2O = (0.182 moles) * (392.1 g/mol) = 71.35 grams
Therefore, if the reaction gives a 100% yield, you would obtain approximately 71.35 grams of K2[Fe(C2O4)3] . 3H2O.
To calculate the number of grams of product obtained, we need to use the concept of stoichiometry and the given moles of the reactant.
First, let's find the molar mass of K2[Fe(C2O4)3] . 3H2O:
molar mass of K2[Fe(C2O4)3] . 3H2O = 2(39.10 g/mol) + 1(55.85 g/mol) + 3(12.01 g/mol x 2) + 6(16.00 g/mol) + 3(1.01 g/mol x 2)
= 78.20 g/mol + 55.85 g/mol + 72.06 g/mol + 96.00 g/mol + 6.06 g/mol
= 307.17 g/mol
Now we can set up a stoichiometric ratio between the reactant and the product.
The balanced chemical equation for the reaction is:
Fe(NH4)2(SO4)2 . 6H2O + 3K2C2O4 = 2K2[Fe(C2O4)3] . 3H2O + (NH4)2SO4
From the equation, we can see that 1 mole of Fe(NH4)2(SO4)2 . 6H2O reacts to give 2 moles of K2[Fe(C2O4)3] . 3H2O.
Given that there are 0.273 moles of Fe(NH4)2(SO4)2 . 6H2O, we can use the stoichiometric ratio to find the corresponding moles of the product:
moles of K2[Fe(C2O4)3] . 3H2O = 0.273 moles x (2 moles/1 mole)
= 0.546 moles
Finally, we can calculate the mass of the product using the molar mass of K2[Fe(C2O4)3] . 3H2O:
mass of K2[Fe(C2O4)3] . 3H2O = moles of K2[Fe(C2O4)3] . 3H2O x molar mass of K2[Fe(C2O4)3] . 3H2O
= 0.546 moles x 307.17 g/mol
= 167.72 grams
Therefore, if the reaction has a 100% yield, you would obtain 167.72 grams of the product, K2[Fe(C2O4)3] . 3H2O.