A network.. where one thing ends another begins
Qualitative properties are properties that are observed and can generally not be measured with a numerical result. They are contrasted to quantitative properties which have numerical characteristics.
Compound 1: Sodium borohydride
In sodium borohydride (NaBH4), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp3 hybridization in NaBH4, to generate 4 hybrid orbitals. These hybrid orbitals, forms sigma bond with 4 'H' atoms. Due to this, the structure of sodium borohydride in tetrahedral.
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Compound 2: B<span>oron trifluoride
</span>In boron trifluoride (BF3), B is a central metal.
Electronic configuration of B is 1s2 2s2 2p1.
B undergoes sp2 hybridization in NaBH4, to generate 3 hybrid orbitals. These hybrid orbitals, forms sigma bond with 3 'H' atoms. Due to this, the structure of <span>boron trifluoride</span> is <span>triangular planner</span>.
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Answer:</h3>
Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.
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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.
