How many liters of chlorine gas, at standard temperature and pressure, will react with 15.5 grams of potassium metal? show all of the work used to solve this problem. 2k + cl2 yields 2kcl

2K +Cl₂ --> 2KCl

First calculate how many moles we have of potassium.

15.5 g * (mol/39.0983 g) = 0.396 mol K

2 moles of K reacts with 1 mole of Cl₂, so we know we'll need 0.198 moles of Cl₂.

Standard temperature and pressure is 273.15 degrees K and 1 atm.

Using the gas law PV = nRT, with R = 0.0821 L*atm*K^-1 *mol^-1, you should be able to solve for V, volume, in Liters.

Sorry i dont understand the question can you rephase it?