Chlorine gas is passed through dissolved potassium iodide until 12.6 g of iodine are precipitated. Dissolved potassium chloride is also produced.

Question

What volume of chlorine was used in this process

Answer 1

Roo,

In questions of this type, first you need to "get" the chemistry involved, then you need to set up your equation and solve it.

For the chemistry, you need to recognize that Cl and I are both halogens, and the one is substituting for the other in the reaction that takes place.

In particular, Cl gas is Cl₂ and potassium iodide is K⁺ and I^- , so that:

Cl₂ + 2 KI -> I₂ + 2 KCl

So what do you do next?

You need a VOLUME of Cl₂ gas. You should know by now, that that translates into a number of MOLES of Cl₂ gas.

So how do you get there?

The key idea here is that everything that happens in a chemical equation happens in terms of MOLES. So you need to go from mass of I₂ to moles of I₂ (divide by molar mass of I₂ for doing that), then transfer that number of moles across into number of moles of Cl₂ using the stoichiometry of the equation -- that's the coefficients on each of the materials in the reaction equation. In this case, there's a coefficient of 1 for both the Cl₂ and the I₂ , so that's a 1<-1 (i.e., 1 mole of Cl₂ was used to make each mole of I₂ produced, but you're tracing back from products to reactants to solve the problem) relationship.

So: first convert mass I₂ to moles I₂. Then recognize that this is the same value as moles Cl₂. Finally, convert from moles of Cl₂ to volume (at STP conditions, unless your problem states otherwise) of Cl₂ gas, using the universal gas law PV = nRT where P is pressure (such as in bar or atm), V is volume (such as in L), n is number of moles of the species, R is the universal gas constant (expressed in whatever units you're using), and T is temperature in K (STP = 273.15 K).

Note also that the standard conditions for two official organizations -- NIST  and IUPAC -- differ a bit on the standard pressure -- the former uses 1 atm, and the latter uses 1 bar ~ 0.987 atm. So make sure that you use whichever calculation, and accompanying value for the universal gas constant, that YOUR course requires!

Hope this helps you; please feel free to provide feedback!