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mihalych1998 [28]
3 years ago
11

LIMITING REACTANT!! Please help I’m very confused.

Chemistry
1 answer:
alexira [117]3 years ago
8 0

Answer:

We'll have 1 mol Al2O3 and 3 moles H2

Explanation:

Step 1: data given

Numer of moles of aluminium = 2 moles

Number of moles of H2O = 6 moles

Step 2: The balanced equation

2Al + 3H2O → Al2O3 + 3H2

Step 3: Calculate the limiting reactant

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

Aluminium is the limiting reactant. It will completely be consumed (2 moles).

H2O is in excess. There will react 3/2 * 2 = 3 moles

There will remain 6 - 3 = 3 moles

Step 4: Calculate moles products

For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2

For 2 moles Al we'll have 2/1 = 1 mol Al2O3

For 2 moles Al We'll have 3/2 * 2 = 3 moles H2

We'll have 1 mol Al2O3 and 3 moles H2

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Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

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Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

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