Question #e14b4

Look at the balanced chemical equation that describes this synthesis reaction

#2"K"_ ((s)) + "Cl"_ (2(g)) -> 2"KCl"_ ((s))#

Notice that it takes #2# moles of potassium and #1# mole of chlorine gas to produce #2# moles of potassium chloride.

The problem doesn't provide any information about the amount of potassium available, which means that you can assume that potassium is in excess, i.e. you have enough potassium around to make sure that all the moles of chlorine gas.

To convert the grams of chlorine gas to moles, use the molar mass of chlorine gas

#1.0 color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "0.0141 moles Cl"_2#

You can use the #1:2# mole ratio that exists between chlorine gas and potassium chloride to say that the reaction will produce

#0.0141 color(red)(cancel(color(black)("moles Cl"_2))) * "2 moles KCl"/(1color(red)(cancel(color(black)("mole Cl"_2)))) = "0.0282 moles KCl"#

To convert this to grams, use the molar mass of potassium chloride

#0.0282 color(red)(cancel(color(black)("moles KCl"))) * "74.5513 g"/(1color(red)(cancel(color(black)("mole KCl")))) = color(darkgreen)(ul(color(black)("2.1 g")))#

The answer is rounded to two sig figs, the number of sig figs you have for the mass of chlorine gas.