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)

1.95 g KClO3

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

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.