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If 90.0 grams of sodium is dropped into 80.0 grams of H2O, how many liters of hydrogen at STP would be produced? Which reactant is in excess and...

Sodium reacts with water to form sodium hydroxide and hydrogen gas. The well balanced chemical equation for this reaction can be written as:


`2Na (s) + 2H_2O -> 2NaOH (aq) + H_2 (g)`


Using stochiometry, 2 moles of sodium reacts with 2 moles of water to form 1 mole of hydrogen gas and 2 moles of sodium hydroxide.


Here, we have 90 g of sodium and 80 g of water. The molar masses of these two species are 23 g and 18 g.


Moles of sodium = 90 g / 23 g/mole = 3.913 moles


moles of water = 80 g / 18 g/mole = 4.444 moles


Since 2 moles of sodium reacts with 2 moles of water or 1 mole of sodium reacts with 1 mole of water, 3.913 moles of sodium will react with 3.913 moles of water and hence water is in excess.


Left over amount of water = 4.444 - 3.913 mole = 0.531 moles 


= 0.531 moles x 18 g/mole = 9.56 g of water.


Thus, 9.56 g of water will be let over.


Since 2 moles of sodium produces 1 mole of hydrogen, 3.913 moles will produce 3.913 x 1/2 moles = 1.96 moles of hydrogen.


At STP, 1 mole of a gas occupies 22.4 l of volume. Thus, 1.96 moles of hydrogen will occupy 1.96 x 22.4 l = 43.9 l.


Hope this helps. 

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