Question

How many grams of H2 can be produced from the reaction of 11.50 g of sodium with an excess of water? The equation for the reaction is 2 Na + 2 H2O -> 2 NaOH + H2 Ans: 0.504 2 g H2 I would like to know how to solve this problem, the teacher gave me the answer but I am unsure how to solve it thanks!

Answer 1

Answer:

$m_{H_2}=0.504gH_2$

Explanation:

Hello,

In this case, since the reaction is:

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

We notice that since there is an excess of water, we can directly compute the yielded grams of hydrogen by using the following stoichiometric procedure, considering the 2:1 molar ratio between sodium and hydrogen (notice the 2 before the sodium and the 1 before the hydrogen at the chemical reaction) and that gaseous hydrogen has a molar mass of 2 g/mol:

$m_{H_2}=11.50gNa*\frac{1molNa}{22.98gNa} *\frac{1molH_2}{2.02molNa} *\frac{2gH_2}{1molH_2} \\\\m_{H_2}=0.504gH_2$

Best regards.

Answer 2

Answer:

0.503g of H₂ are produced

Explanation:

Based on the chemical reaction:

2 Na + 2 H₂O → 2 NaOH + H₂

2 moles of Na react with 2 moles of water to produce 2 moles of NaOH and 1 mole of H₂

11.50g of Na -limiting reactant, molar mass 22.99g/mol- are:

11.50g× (1mol / 22.99g) = 0.500 moles of Na.

As 2 moles of Na produce 1 mole of H₂:

0.500 moles of Na × (1 mole H₂ / 2 moles Na) = 0.250 moles of H₂

As molar mass of H₂ is 2.01g/mol:

0.250 moles of H₂ × (2.01g / 1mol) = 0.503g of H₂ are produced