The volume of fluorine gas required to react with 2.67 g of calcium bromide to form calcium fluoride and bromine at 41.0 °C and 4.31 atm is ________ mL.
A)10.4
B)210
C)420
D)79.9
E)104

Question

Quizplus · Accepted Answer

1) Write the balanced chemical equation:

CaBr2 + F2 → CaF2 + 2Br2

2) Calculate the number of moles of CaBr2:

m = 2.67 g 
M = 199.89 g/mol 
n = m/M = 0.0134 mol

3) Use stoichiometry to calculate the number of moles of F2 required:

From the balanced chemical equation, 1 mole of CaBr2 reacts with 1 mole of F2.

Therefore, 0.0134 mol of CaBr2 will react with 0.0134 mol of F2.

4) Use the ideal gas law to calculate the volume of F2 required:

PV = nRT

V = nRT/P

V = (0.0134 mol) (0.0821 L·atm/K·mol) (314.15 K) / (4.31 atm)

V = 0.79 L = 790 mL

Therefore, the volume of F2 required is 79.9 mL (rounded to three significant figures).

The answer is D.

The Volume of Fluorine Gas Required to React with 2