Question #b7520

Before doing anything else, start by converting the masses of the two reactants to moles by using their respective molar masses

#122.6 color(red)(cancel(color(black)("g"))) * "1 mole XeF"_6/(245.28color(red)(cancel(color(black)("g")))) = "0.4986 moles XeF"_6#

#60 color(red)(cancel(color(black)("g"))) * "1 mole SiO"_2/(60.08color(red)(cancel(color(black)("g")))) = "0.9987 moles SiO"_2#

Now, the balanced chemical equation

#2"XeF"_ (6(s)) + "SiO"_ (2(s)) -> 2"XeOF"_ (4(l)) + "SiF"_ (4(g))#

tells you that the reaction consumes #2# moles of xenon hexafluoride for every #1# mole of silicon dioxide.

At this point, it should be clear that xenon hexafluoride is the limiting reagent because, in order for all the moles of silicon dioxide to react, you will need

#0.9987 color(red)(cancel(color(black)("moles SiO"_2))) * "2 moles XeF"_6/(1color(red)(cancel(color(black)("mole SiO"_2)))) = "1.9974 moles XeF"_6#

Since you only have #0.4986# moles of xenon hexafluoride, you can say that this compound will act as the limiting reagent, i.e. it will be completely consumed before all the moles of silicon dioxide will get the chance to react.

Now, you also know that it takes #2# moles of xenon hexafluoride to produce #1# mole of silicon tetrafluoride, so you can say that the reaction will produce

#0.4986 color(red)(cancel(color(black)("moles XeF"_6))) * "1 mole SiF"_4/(2color(red)(cancel(color(black)("moles XeF"_6)))) = "0.2493 moles SiF"_4#

To convert this to grams, use the molar mass of silicon tetrafluoride

#0.2493 color(red)(cancel(color(black)("moles SiF"_4))) * "104.08 g"/(1color(red)(cancel(color(black)("mole SiF"_4)))) = "26 g"#

So the correct answer would be (1) #"XeF"_6 and "26 g"#.