A 2.60 g chunk of sodium is added to a flask of solvent contaminated with 1.25 g

Question

A 2.60 g chunk of sodium is added to a flask of solvent contaminated with 1.25 g of water. In order to calculate the mass of sodium left after the reaction has completed, we must first find the number of moles of sodium left after the reaction has completed: mNa-left = mNa-start+Na-reacted Based on the balanced chemical equation, we know that 2 moles of Na react with every 2 moles of water (which we have already determined is the limiting reagent). So, we can: Calculate the number of moles of water reacted: Calculate the number of moles of sodium reacted: Calculate the number of moles of sodium left:

Explanation / Answer

2 Na (s) + H20(l)---------------------> 2 NaOH(aq)

46 grams of sodium react with   ------------------------18 grams of water

2.60 grams of sodium reacts with---------------------------------------------?

=2.60*18/46

=1.017 grams of water

no.of mole of water reacted= 1.017/18 =0.0565 moles of water

therefore limiting reagent is sodium

no.of moles of sodium reacted=2.60/23=0.113 moles of sodium

the amount of water left = 0.232 grams

the amount of sodium left=0

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