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What is the limiting reagent when 22 g of sodium is added to 28 g water?

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Answer:

Limiting reagent is Na

Explanation:

Reaction of Na with water is:

Na + H₂O →  NaOH  +  H₂

It is a very exothermic reaction.

The correctly balanced equation is:

2Na + 2H₂O →  2NaOH  +  H₂

Ratio is 2:2, between reactants, according to stoichiometry.

We convert mass to moles:

22 g . 1 mol / 23 g = 0.956 moles of Na

28 g . 1mol / 18 g = 1.56 moles of water.

Certainly the limiting reagent is sodium. For 1.56 moles, we need the same amount of sodium and we only have 0.956 moles.

We do not have enough sodium to complete the reaction.

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