Oxygen can be generated by the thermal decompisition of Potassium Chlorate, desc

Question

Oxygen can be generated by the thermal decompisition of Potassium Chlorate, described in the following equation: 2KClO3(s)---> 2KCl(s) + 3O2(g)

a. Calculate the volume of oxygen,measured at STP, produced by the decompistion of 2.87 g of potassium chlorate.

b. Calculate the volume (in mL) measur3ed at 27 degrees C and 712 torr, produced by the decompisition of 3.904 g of potassium chlorate.

Thanks

Explanation / Answer

2KClO3(s)---> 2KCl(s) + 3O2(g) M.W=122.55 i.e 122.55 grams of KClO3 gives (3/2)*32g=(3/2)*22.4l of oxygen gas at STP so 2.87 g will give: 2.87*( (3/2)*22.4l)/122.55l=0.786l of oxygen similarly 3.904 g will give 1.07l at stp by boyles law P1V1=P2V2 P1=760, P2=712 V1=1.07,V2=? V2=1.14l

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