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qaws [65]
2 years ago
5

Determine the limiting reactant in each of the following reactions:

Chemistry
1 answer:
wolverine [178]2 years ago
3 0

The reactant in a chemical process known as the limiting reactant controls how much product can be produced. When the limiting reactant is completely used up, the reaction will come to an end.

<h3>Find the limiting reactant ?</h3>
  • As a result of 1 mol Sb4O6 reacting with 6 mol H2SO4, only 0.1 mol Sb4O6 reacts with 0.6 mol H2SO4, leaving only 0.5 mol H2SO4. This indicates that H2SO4 is the limiting reactant and Sb4O6 is present in excess.
  • According to your equation, which is balanced, 0.1 mol Sb4O6 should react with 0.6 mol H2SO4, yet there is only 0.5 mol H2SO4 on hand.
  • Therefore, only.083 mol of Sb4O6 are reacted.
  • The reactant that is present in the limiting amount—the limiting reactant—determines the extent to which a chemical reaction occurs.
  • The trick is really quite easy! We employ an augmented matrix to hold the data derived from the balancing equation Sb4O6 + 6H2SO4 --> 2Sb2(SO4)3 + 6H2O.
  • Although you are provided 0.5 mol of H2SO4, the reaction requires 0.6 mol. Therefore, the limiting reactant is H2SO4.
  • Only 0.0833 mol of Sb4O6 is required, but you have 0.1 mol. Sb4O6 is therefore the extra reactant.

To learn more about limiting reactant refer to:

brainly.com/question/27986321

#SPJ1

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How much aluminum oxide are produced when 46.5g of Al react with 165.37g of MnO?
solong [7]

Aluminum oxide produced : = 79.152 g

<h3>Further explanation</h3>

Given

46.5g of Al

165.37g of MnO

Required

Aluminum oxide produced

Solution

Reaction

2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)

  • mol Al(Ar = 27 g/mol) :

mol = mass : Ar

mol = 46.5 : 27

mol = 1.722

  • mol MnO(Ar=71 g/mol) :

mol = 165.37 : 71

mol = 2.329

mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776

MnO as a limiting reactant(smaller ratio)

So mol Al₂O₃ based on MnO as a limiting reactant

From equation , mol Al₂O₃ :

= 1/3 x mol MnO

= 1/3 x 2.329

= 0.776

Mass Al₂O₃ (MW=102 g/mol) :

= 0.776 x 102

= 79.152 g

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