The percent potassium chlorate in the unknown generated in the reaction is known

Question

The percent potassium chlorate in the unknown generated in the reaction is known. In this procedure there are two ways to calculate the moles of oxygen: 1) by using the weight of oxygen (see IA) and 2) by using the volume of oxygen generated at STP (see IB) and the accepted value for molar volume. Using both values for the number of moles of oxygen generated, answer the following questions, showing all calculations and the proper number of significant figures. Refer to the balanced equation for the reaction in order to go from moles of oxygen formed to moles of potassium chlorate decomposed. a. the number of moles of oxygen formed b. the number of moles of potassium chlorate which decomposed to form this amount of oxygen c. the number of grams of potassium chlorate present in the sample d. the percent potassium chlorate in the sample

Explanation / Answer

mol of O2 = mass/MW = 0.53/32 = 0.0165625 mol of O2

ratio is:

3 mol of O2 --> 2 mol of KClO3

so

0.0165625 mol --> 2/3*0.0165625 = 0.011041 mol of KClO3

mass of KClO3 in sample --> mol*MW = 0.011041*122.55 = 1.3530 g of KClO3

d)

% KClO3 in sample -->

% mass --> Mass of KClO3 / total mass * 100%

we need total mass of sample... but simpley substitute your value in :

% mass --> 1.3530   / total mass * 100%

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