How much H2 gas at STP can be produced by
the reaction
2 Na(s) + 2 H2O(ℓ)heat−→ H2(g) + 2 NaOH(aq)
of 3.90 g of Na and excess water?
Answer in units of L.
mols Na = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Na to mols H2.
Then mols H2 x (22.4 L/mol) = ?L H2 gas @ STP.
I need the answer so I can check my work.
To find out the amount of H2 gas produced at STP, we need to use the given information and some stoichiometry calculations.
First, let's find the number of moles of sodium (Na) using its molar mass. The molar mass of Na is 22.99 g/mol.
Number of moles of Na = Mass of Na / Molar mass of Na
Number of moles of Na = 3.90 g / 22.99 g/mol
Number of moles of Na = 0.170 mol
According to the balanced chemical equation, the ratio of Na to H2 is 2:1. This means that for every 2 moles of Na, 1 mole of H2 is produced.
Number of moles of H2 = Number of moles of Na / 2
Number of moles of H2 = 0.170 mol / 2
Number of moles of H2 = 0.085 mol
Now, we can use the ideal gas law to find the volume of the H2 gas produced at STP. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (STP = 273 K)
Rearranging the equation to solve for volume:
V = nRT / P
Substituting the values into the equation:
V = (0.085 mol) * (0.0821 L·atm/(mol·K) * (273 K) / (1 atm)
V ≈ 1.76 L
Therefore, approximately 1.76 liters of H2 gas can be produced at STP by the given reaction.