5F248+ 2NH3(8) ► NF418 + 6HF a) If you start with 3.6 g of NH, and excess F2, how many liters of HF at STP can you make?

14.2L at STP

Explanation:

Based on the problem, 2 moles of NH3 produce 6 moles of HF. To solve this question we have to convert the mass of NH3 to moles. With the chemical equation find the moles of HF and using PV = nRT find the liters of HF:

Moles NH3 -Molar mass: 17.031g/mol-

3.6g NH3 * (1mol / 17.031g) = 0.211 moles NH3

Moles HF:

0.211 moles NH3 * (6mol HF / 2mol NH3) = 0.634 moles HF

Volume HF

PV = nRT; V = nRT/P

Where V is volume in liters, n are moles of the gas = 0.634 moles, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP and P is pressure = 1atm at STP.

Replacing:

V = 0.634moles*0.082atmL/molK*273.15K / 1atm

V = 14.2L at STP