Question

CHM-120 Lab 7 Percent Composition of KCIO; Rev2-17 Page 2 of 10 Prelab, cont. 3. In both question 1 and 2 above explain why you retain less than half of the coins but the value and mass of what remains is greater than half of what you started with If 0.3785 grams of calcium was completely reacted with oxygen gas in the air to form calcium oxide, what mass of calcium oxide would be expected? The reaction is as follows: 2C +02 2Cao (note molecular oxygen is O, as shown in the equation) 5. Calculate the number of moles of potassium chlorate (KCIO) in a sample of potassium chlorate that has a mass of 1.384 grams.

Answer 1

4. Ans :-

Given mass of Ca = 0.3785 g

Gram molar mass of Ca = 40.078 g/mol

Because,

Number of moles = Given mass of substance/Gram molar mass of substance

So,

Number of moles of Ca = Given mass of Ca/Gram molar mass of Ca

= 0.3785 g / 40.078 g/mol

= 0.00944 mol

Given, Chemical equation is : 2 Ca (s) + O₂ (g) ------------> 2 CaO (s)

It is clear that,

2 mol of Ca gives = 2 mol of CaO

So,

0.00944 mol of Ca gives = 0.00944 mol of CaO

Now,

Mass of CaO = Number of moles x Gram molar mass of CaO

= 0.00944 mol x 56.0774 g/mol

= 0.5294 g

Hence, mass of CaO formed = 0.5294 g

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5.Ans :-

Number of moles of KClO₃ = Given mass of KClO₃ / Gram molar mass of KClO₃

= 1.384 mol / 122.55 g/mol

= 0.0113 mol

Hence, Number of moles of KClO3 = 0.0113 mol