Final answer:
To find out how many moles of sodium oxide (Na2O) can be formed when 21.7 mL of sodium (Na) reacts with excess oxygen gas, we need to use the given density of sodium (0.97 g/mL). First, convert the volume of sodium to grams by multiplying the volume by the density. Next, convert the mass of sodium to moles by dividing the mass by the molar mass of sodium. Finally, use the balanced equation to calculate the moles of sodium oxide formed based on the moles of sodium.
Step-by-step explanation:
To find out how many moles of sodium oxide (Na2O) can be formed when 21.7 ml of sodium (Na) reacts with excess oxygen gas, we need to use the given density of sodium (0.97 g/mL).
First, we can convert the volume of sodium to grams by multiplying the volume (in mL) by the density (in g/mL). This gives us:
21.7 mL Na x 0.97 g/mL = 21.1 g Na
Next, we can convert the mass of sodium to moles by dividing the mass by the molar mass of sodium (22.99 g/mol). This gives us:
21.1 g Na / 22.99 g/mol = 0.917 mol Na
Now, we can use the balanced equation (which is Na + O2 -> Na2O) to determine the moles of sodium oxide formed. According to the equation, 2 moles of sodium react with 1 mole of oxygen gas to produce 2 moles of sodium oxide. Therefore, since we have 0.917 mol Na, we can calculate the moles of sodium oxide as:
0.917 mol Na x (2 mol Na2O / 2 mol Na) = 0.917 mol Na2O
So, the correct answer is 0.917 mol Na2O.
To find the moles of sodium oxide formed, first calculate the mass of sodium that reacts, then convert the mass to moles. Using the molar stoichiometry from the balanced chemical equation, find the proportion of moles of sodium oxide produced. The correct answer is 0.458 moles of sodium oxide.
The question asks how many moles of sodium oxide (Na2O) can be formed when 21.7 ml of sodium (Na) with a density of 0.97 g/mL reacts with excess oxygen gas. The first step in solving this is to calculate the mass of sodium that reacts:
Mass of Na = Volume × Density = 21.7 mL × 0.97 g/mL = 21.049 g
Next, we convert the mass of sodium to moles by using its molar mass (22.99 g/mol):
Moles of Na = Mass of Na / Molar Mass of Na = 21.049 g / 22.99 g/mol ≈ 0.915 moles Na
The balanced chemical equation for the reaction of sodium with oxygen to form sodium oxide is:
4Na (s) + O2 (g) → 2Na2O (s)
From the stoichiometry of the balanced equation, 4 moles of Na produce 2 moles of Na2O. By setting up a simple proportion, we can find the moles of Na2O formed:
(0.915 moles Na) × (2 moles Na2O / 4 moles Na) = 0.4575 moles Na2O
Therefore, option D, 0.458 moles Na2O, is the correct answer.