If .275 moles of ferrous ammonium sulfate (Fe(NH4)2(SO4)2. 6H2O) and an excess of all other reagents are used in a synthesis of K3[Fe(C2O4)3].3H2O, how many grams of product will be obtained if the reaction gives a 100% yield?
Can someone help me solve this I found the molar mass of (Fe(NH4)2(SO4)2. 6H2O) = 392.1g/mol and K3[Fe(C2O4)3].3H2O=491.1 g/mol
Thank You
Question...
So then is it unnecessary to calculate the molar mass of the reagent.?
Is only the molar mass of the product relevant to this question.?
To find the mass of the product, K3[Fe(C2O4)3]·3H2O, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that runs out first and determines the amount of product that can be formed.
First, let's calculate the number of moles of K3[Fe(C2O4)3]·3H2O that can be formed from the given amount of ferrous ammonium sulfate (Fe(NH4)2(SO4)2·6H2O).
1. Calculate the molar mass of ferrous ammonium sulfate (Fe(NH4)2(SO4)2·6H2O):
Molar mass = ([(1x2) + (14x2) + (1x8)] x 2) + (32x1) + (16x8) + (6x2) = 392.1 g/mol
2. Calculate the number of moles of ferrous ammonium sulfate (Fe(NH4)2(SO4)2·6H2O):
moles = mass / molar mass = 0.275 moles
3. The balanced equation for the reaction is:
Fe(NH4)2(SO4)2·6H2O + 3K2C2O4·H2O → K3[Fe(C2O4)3]·3H2O + (NH4)2SO4 + 2K2SO4 + 8H2O
4. From the balanced equation, the stoichiometry between Fe(NH4)2(SO4)2·6H2O and K3[Fe(C2O4)3]·3H2O is 1:1. This means that 1 mole of Fe(NH4)2(SO4)2·6H2O reacts to produce 1 mole of K3[Fe(C2O4)3]·3H2O.
5. Therefore, the number of moles of K3[Fe(C2O4)3]·3H2O formed is also 0.275 moles.
6. Calculate the mass of K3[Fe(C2O4)3]·3H2O:
mass = moles x molar mass = 0.275 moles x 491.1 g/mol = 135.105 g
Therefore, if the reaction proceeds with a 100% yield, the mass of K3[Fe(C2O4)3]·3H2O obtained will be 135.105 grams.
To calculate the mass of the product (K3[Fe(C2O4)3].3H2O) formed, we need to use the stoichiometry of the balanced equation.
The balanced equation for the reaction is:
Fe(NH4)2(SO4)2.6H2O + 3K2C2O4 + 3H2C2O4 -> K3[Fe(C2O4)3].3H2O + (NH4)2SO4 + 3H2O
From the balanced equation, we can see that 1 mole of Fe(NH4)2(SO4)2.6H2O reacts with 3 moles of K2C2O4 and 3 moles of H2C2O4 to produce 1 mole of K3[Fe(C2O4)3].3H2O.
First, let's calculate the moles of Fe(NH4)2(SO4)2.6H2O given:
Moles = mass / molar mass
Moles = 0.275 moles (given)
Now, using the stoichiometry of the balanced equation, we can determine the moles of K3[Fe(C2O4)3].3H2O produced:
Moles of K3[Fe(C2O4)3].3H2O = 0.275 moles * (1 mole K3[Fe(C2O4)3].3H2O / 1 mole Fe(NH4)2(SO4)2.6H2O)
Moles of K3[Fe(C2O4)3].3H2O = 0.275 moles
Finally, we can convert the moles of K3[Fe(C2O4)3].3H2O to grams:
Mass = moles * molar mass
Mass = 0.275 moles * 491.1 g/mol
Mass ≈ 135.05 grams
Therefore, if the reaction goes to 100% yield, approximately 135.05 grams of K3[Fe(C2O4)3].3H2O will be obtained.
since they each have 1 Fe, 1 mole of the Fe(NH4)2(SO4)2· 6H2O makes 1 mole of the K3[Fe(C2O4)3]·3H2O
so, since you start with .275 moles of (Fe(NH4)2(SO4)2. 6H2O) you end up with .275 moles of K3[Fe(C2O4)3].3H2O.
Now change that to grams and you're done.
do a web search for your question, and you'll find the problem answered.