What volume of hydrogen gas can be produced by reacting 4.20 g of sodium in excess water at 50.0 celsius and 106 kpa? the reaction is 2na + 2h20 --> 2naoh + h2.

From the equation, you can conclude that 2 Natrium atoms will form 1 H2 molecule. Then, the number of hydrogen molecules formed by 4.2g of sodium would be:4.2g /(23g/mol) * 1/2= 0.0913 mol. 

Using the ideal gas formula, the volume of the gas would be:
V= nRT/P
V= (0.0913 mol * 8.314 L-kPa/mol-K *323K )/ 106 kPa
V= 2.3131 L

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