Answer:
9.07 L
Explanation:
Calculate the volume occupied at s.t.p by 6.89 g of NH₃ gas [H = 1.0, N = 14.0].
Step 1: Given and required data
- Molar mass of NH₃ (M): 17.0 g/mol
Step 2: Calculate the moles (n) of NH₃
We will use the following epxression.
n = m / M
n = 6.89 g / (17.0 g/mol) = 0.405 mol
Step 3: Calculate the volume occupied by 0.405 moles of NH₃ at STP
At STP, 1 mole of NH₃ occupies 22.4 L (assuming ideal behavior).
0.405 mol × 22.4 L/1 mol = 9.07 L
