What volume of chlorine gas is required to react completely with 0.390 mol of iron according to the following reaction at 0°C and 1 atm?

Question

What volume of chlorine gas is required to react completely with 0.390 mol of iron according to the following reaction at 0°C and 1 atm?

iron(s) + chlorine(g) --> iron(ii) chloride(s)

Volume = _____L

Answer 1

This problem requires you to use to equation for the ideal gas law, PV=nRT, where P is pressure, V is volume, n is moles of the gas, R is the gas constant, and T is temperature.

This equation can be rearranged to solve for V by dividing both sides of the equation by P. Thus, V=(nRT)/P

Next, we must use the moles of iron (given) to calculate the moles of chlorine gas. We can do this by first balancing the equation then using mole ratios.

Unbalanced equation: Fe (s) + Cl2 (g) --> FeCl3 (s)

Balanced equation: 2Fe (s) + 3Cl2 (g) --> 2FeCl3 (s)

Now we know that this reaction consumes 3 moles of Cl2 (g) per 2 moles of Fe (s).

So, 0.390 mol Fe (s) * ((3 mol Cl2)/(2 mol Fe)) = 0.585 mol Cl2 (g)

Now we have everything that we need to solve for V.

n= 0.585 mol Cl2 (g) (we just calculated this value)

R= 0.082 (L*atm)/(K*mol) (this is a constant value)

T= 0 degrees C= 273 K (given in the question, then converted to K)

P= 1 atm (given in the question)

V= (nRT)/P = (0.585*0.082*273)/1 = 13.1 L