If i perform this reaction with 25grams of iron (iii) phosphate and an excess of sodium sulfate how many grams of iron (iii) sulfate can i make?
Answers
Answer:
So, the first step in any stoichiometry question is to find the molecular weights of the compounds involved:
FePO4 = 55.845+ 30.974 + 16*4 = 150.819 g/mol
Fe2(SO4)3 = 55.845*2 + (32.065+16*4)*3= 399.885 g/mol
The next step is to find the number of moles of your reactant, FePO4:
grams of substance/molecular weight of substance= moles of substance
25g/150.819g/mol= 0.166 moles of FePO4
Then, think about the proportion of how many moles of your reactant there are for each mole of your product. In this case, there are 2 moles of FePO4 for every 1 mole of Fe2(SO4)3.
0.166 mole FePO4 * 1 mole Fe2(SO4)3/ 2 moles of FePO4 = .083 mole of Fe2(SO4)3
Now, multiply the number of moles of your product by its molecular weight:
0.083 moles of Fe2(SO4)3 * 399.885 g/mol = 33.190 grams of Fe2(SO4)3
Let me know if you have any further questions.
Explanation:
25 grams of FePO4 will produce 66.36 grams of Fe2(SO4)3.
To answer this question, we need to write and balance the chemical equation for the reaction between iron (III) phosphate and sodium sulfate:
FePO4 + 3 Na2SO4 → 3 Na3PO4 + Fe2(SO4)3
According to the balanced equation, for every 1 mole of FePO4, we can obtain 1 mole of Fe2(SO4)3.
We first need to convert the given amount of FePO4 into moles:
molar mass of FePO4 = 150.82 g/mol
moles of FePO4 = mass / molar mass = 25 g / 150.82 g/mol = 0.166 moles
Since sodium sulfate is present in excess, it means that it will not be limiting the reaction.
Therefore, all of the 0.166 moles of FePO4 will react to form an equal number of moles of Fe2(SO4)3.
The molar mass of Fe2(SO4)3 is 399.88 g/mol. Thus, the mass of Fe2(SO4)3 produced can be calculated as:
mass of Fe2(SO4)3 = moles of Fe2(SO4)3 × molar mass
= 0.166 moles × 399.88 g/mol
= 66.36 g
Therefore, 25 grams of FePO4 will produce 66.36 grams of Fe2(SO4)3.
For such more questions on reaction,
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