Question

Calculate the volume in liters of hydrogen gas produced when 5.00 grams of sodium metal is placed in 12.50 g of water at 31.5°C and 888 Torr. 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)

Answer 1

The volume of hydrogen gas produced when 5.00 grams of sodium reacts with water at 31.5°C and 888 Torr is 2.425 liters, calculated using stoichiometry and the ideal gas law.

Step-by-step explanation:

To calculate the volume of hydrogen gas produced when 5.00 grams of sodium metal reacts with water, we'll use the stoichiometry of the reaction and the ideal gas law (PV = nRT).

First, we calculate the moles of sodium:

5.00 g Na * (1 mol Na / 22.99 g Na) = 0.217 moles Na

According to the balanced chemical equation, 2 moles of Na produce 1 mole of H2 gas, so we will have 0.1085 moles of hydrogen gas.

Next, we use the ideal gas law to find the volume:

V = (nRT) / P

Where:

R = 0.0821 L atm/mol K (ideal gas constant)

T = 31.5°C + 273.15 = 304.65 K

P = 888 Torr * (1 atm / 760 Torr)

= 1.169 atm

V = (0.1085 mol * 0.0821 L atm/mol K * 304.65 K) / 1.169 atm

= 2.425 L

Therefore, the volume of hydrogen gas produced under the given conditions is 2.425 liters.