Answer to Question #73638 in General Chemistry for Madeline

Question

if you react 16.3g of SCl2 with 10.5g of NaF, what is the maximum amount of SF4 that can be produced? Equation: 3SCl2 + 4NaF ---> SF4 + S2Cl2 + 4NaCl FW: (103.1) (42) ---> (108) (135.2) (58.5)

Answer 1

First we have to calculate limiting reagent
(reagent that determines the amount of product that can be formed by a reaction; the reaction occur only until the limiting reagent is used up)
n(SCl2) = m(SCl2)/M(SCl2)=16.3/3*103.1=0.053 mol
n(NaF) = m(NaF)/M(NaF)=10.5/4*42=0.062 mol
SCl2 is limiting reagent.
Then m(SF4) = 16.3*108/4*103.1=5.69 g.