<h3>
Answer:</h3>
Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.
<h3>
Explanation:</h3>
From the questions we are given;
4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(s)
- Moles of FeS₂ are 26.62 moles
- Moles of Oxygen, O₂ are 59.44 moles
We are supposed to determine the limiting and excess reactants;
- From the equation of the reaction given; 4 moles of FeS₂ required 11 moles of Oxygen gas.
Working with the amount of reactants given;
- 26.62 moles of FeS₂ will require 73.205 moles of O₂ and only 59.44 moles of O₂ are available.
- On the other hand 59.44 moles of O₂ requires 21.615 moles of FeS₂, and we are given 26.62 moles of FeS₂ which means FeS₂ is in excess.
Conclusion;
We can conclude that Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.
Words that includes names, places, geometric shapes and finally units of measure.
Answer:
0.375 grams are needed to make 25 mL solution.
Explanation:
Mass of 
cuprous nitrate required to make 1 l of solution = 15 g.
1 L = 1000 mL
Mass of cuprous nitrate required to make 1000 mL of solution = 15 g
Mass of cuprous nitrate required to make 1 mL of solution:
Mass of cuprous nitrate required to make 25 mL of solution:
0.375 grams are needed to make 25 mL solution.
Answer:
the 2 means there are 2 al and the 3 means there are 3 o
Explanation:
I hope this helps
Answer:We are already given with the mass of the Xe and it is 5.08 g. We can calculate for the mass of the fluorine in the compound by subtracting the mass of xenon from the mass of the compound.
mass of Xenon (Xe) = 5.08 g
mass of Fluorine (F) = 9.49 g - 5.08 g = 4.41 g
Determine the number of moles of each of the element in the compound.
moles of Xenon (Xe) = (5.08 g)(1 mol Xe / 131.29 g of Xe) = 0.0387 mols of Xe
moles of Fluorine (F) = (4.41 g)(1 mol F/ 19 g of F) = 0.232 mols of F
The empirical formula is therefore,
Xe(0.0387)F(0.232)
Dividing the numerical coefficient by the lesser number.
XeF₆
Explanation:
