Question

The equation for the reaction between lithium and water is 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g). A mass of 0.500 g of lithium reacts with an excess of water. Calculate the volume, in cm^3, of hydrogen gas produced at rtp. [molar volume of a gas at rtp = 24 000 cm^3]. Give your answer to three significant figures.

A) 12 cm^3
B) 24 cm^3
C) 48 cm^3
D) 72 cm^3

Answer 1

The volume of hydrogen gas produced at rtp from the reaction of 0.500 g of lithium with water is calculated to be 864 cm³, which does not match the given answer choices.

Step-by-step explanation:

To calculate the volume of hydrogen gas produced from the reaction between lithium and water, we use the stoichiometry of the balanced chemical equation and apply the concept of molar volume at room temperature and pressure (rtp).

Step-by-Step Solution:

1. Write the balanced chemical equation: 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g). This shows that 2 moles of lithium produce 1 mole of hydrogen gas.
2. Calculate the number of moles of lithium used by dividing the mass of lithium by its molar mass: molar mass of Li = 6.941 g/mol.
   Moles of Li = 0.500 g ÷ 6.941 g/mol = 0.072 moles (to three significant figures).
3. Determine the number of moles of hydrogen gas produced using the stoichiometry from the equation: Moles of H2 = 0.072 moles Li × (1 mole H2 / 2 moles Li) = 0.036 moles H2.
4. Calculate the volume of hydrogen gas at rtp using the molar volume: Volume of H2 = 0.036 moles × 24,000 cm³/mol = 864 cm³.
5. Since the volume required is expressed in cm³, no additional conversion is necessary.

The calculated volume of 864 cm³ does not match any of the given answer choices, indicating there may be a typo or misinformation in the question or answer choices.