In the reaction 2 Na + 2 H₂O → 2 NaOH + H₂, how many liters of hydrogen at STP are produced from 50.0 grams of sodium (Na = 23.0, O = 16.0, H = 1.0 g/mol)?

Question

Quizplus · Accepted Answer

First, we need to balance the equation.
2 Na + 2 H₂S → 2 NaOH + H₂S
Now we can use stoichiometry to find the amount of hydrogen produced. 
From the equation, we can see that for every 2 moles of Na, 1 mole of H₂S is produced. Additionally, 1 mole of H₂S produces 1 mole of H₂. 
Using the given mass of sodium (50.0 g), we can convert it to moles:
50.0 g Na x (1 mol Na/23.0 g Na) = 2.17 mol Na
Now we can use stoichiometry:
2.17 mol Na x (1 mol H₂S/2 mol Na) x (1 mol H₂/1 mol H₂S) = 0.542 mol H₂
Finally, we can use the ideal gas law at STP (standard temperature and pressure):
PV = nRT
(1 atm) V = (0.542 mol) (0.08206 L*atm/mol*K) (273 K)
V = 12.3 L
However, the question asks for the volume of hydrogen gas produced, so we need to divide by the stoichiometric coefficient (2) since we have 2 moles of Na:
V = 6.15 L = (50.0/23.0)(22.4/2) L

In the Reaction 2 Na + 2 H2O → 2

In the Reaction 2 Na + 2 H₂O → 2