Question

how many moles would be present in 300 liters of dinitrogen monoxide at stp?
a 12.3 moles
b 12.4 moles
c 13.4 moles
d 6,720 moles

Explanation:

Step1: Recall STP molar volume

At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. So the formula to find moles from volume is $n = \frac{V}{22.4\space L/mol}$, where $n$ is moles and $V$ is volume in liters.

Step2: Substitute volume into formula

Given $V = 300\space L$, substitute into the formula: $n=\frac{300\space L}{22.4\space L/mol}$.
Calculate the value: $\frac{300}{22.4}\approx13.4$ moles? Wait, no, wait, 300 divided by 22.4: 22.413 = 291.2, 300 - 291.2 = 8.8, 8.8/22.4≈0.39, so total≈13.39, close to 13.4? Wait, but let me check again. Wait, maybe I made a mistake. Wait, 22.4 L/mol is the molar volume. So 300 L / 22.4 L/mol = 300/22.4 ≈13.39, which is approximately 13.4 moles. Wait, but the options: c is 13.4 moles. Wait, but let me check the calculation again. 22.4 13 = 291.2, 300 - 291.2 = 8.8, 8.8 /22.4 = 0.392857, so total 13.392857, which is ~13.4 moles. So the correct option is c.

Answer:

c. 13.4 moles