What mass of dihydrogen gas results from the oxidation of #20*g# sodium metal by a mass of #10*g# of water?

Question

Avery Adams · Accepted Answer

Well first we write a stoichiometric equation, and get a bit over #1*g# of #H_2(g)#

The following stoichiometric equation is used to represent the reaction:

#Na(s) + H_2O(l) rarr NaOH(aq) + 1/2H_2(g)uarr#

You can double this if you like. I think the arithmetic is easier if you include the #1/2# coefficient.......

#"Moles of sodium"=(20*g)/(22.99*g*mol^-1)=0.870*mol#

#"Moles of water"=(10*g)/(18.01*g*mol^-1)=0.555*mol#.

Now, here (UNUSUALLY) water is the reagent in DEFICIENCY. And thus we assume that only #0.555*mol# of water reacts, and that some of the sodium metal remains unreacted - an unusual scenario.

And thus we get #(0.555*mol)/2# dihydrogen gas evolved, i.e. #(0.555*mol)/2xx2.02*g*mol^-1=1.11*g#

Natalie Anton · Answer

undefined To determine the mass of dihydrogen gas produced from the oxidation of sodium metal by water, you need to consider the balanced chemical equation for the reaction between sodium and water:

$$2Na + 2H_2O \rightarrow 2NaOH + H_2$$

From the equation, you see that 2 moles of sodium react with 2 moles of water to produce 1 mole of dihydrogen gas.

Now, you need to find the number of moles of sodium and water given the masses provided. Then, you can use stoichiometry to find the mass of dihydrogen gas produced.