Question

How many grams of solid potassium chlorate (KCIO3) decompose to form solid

potassium chloride and 790 mL of oxygen gas at 111 °C and 770 torr? (3 sf)

Answer 1

1.95 g KClO3

Step-by-step explanation:

We're asked to find the mass, in g, of KClO3 that decomposes to give a certain amount of O2.

Let's use the ideal gas equation to find the moles of O2 that form.

P=747torr(1latm760torr)=0.983 atmV=797mL(1lL103mL)=0.797

LT=128oC+273=401 KPlugging in known values, we haven=PVRT=(0.983atm)(0.797L)(0.082057L∙atmmol∙K)(401K)=0.0238 mol O2

Now, we'll use the coefficients of the chemical equation to find the relative number of moles of KClO3 that reacted:0.0238mol O2(2lmol KClO33mol O2)=0.0159 mol KClO3

Finally, we'll use the molar mass of potassium chlorate (122.55 g/mol) to find the number of grams that reacted:

0.0159mol KClO3(122.55lg KClO31mol O2)=1.95 g KClO3