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Question Question If an inflatable device has a volume of 1.57 x 106 L. And in order to use enough O2 to inflate it we use this reaction: 2 KClO3 (s) → 3 O2 (g) + 2 KCl (s) If the temperature is 4.00 C and the pressure is 1.00 atm, how many grams of KClO3 (s) are required to produce 1.57 x 106 L of O2 (g)? R=0.08206 Lxatm/molxk Pv=nrt (1.00atm)(1.57×106 L)=(x)(0.08206 L*atm/mol*k)(277.15) 1.57×106 L = 22.742929(x) x=69032.44521 mol O2 69032.44521 mol O2 * 2 mol KClO3 / 3 mol O2 = 138064.8904 mol KClO3 138064.8904 mol KClO3 * 122.548 g / 1 mol KClO3 16919576.19 g KClO3 Trying to make sure my calculations are right. Thanks. And in order to use enough O2 to inflate it we use this reaction: 2 KClO3 (s) → 3 O2 (g) + 2 KCl (s) If the temperature is 4.00 C and the pressure is 1.00 atm, how many grams of KClO3 (s) are required to produce 1.57 x 106 L of O2 (g)? R=0.08206 Lxatm/molxk Pv=nrt (1.00atm)(1.57×106 L)=(x)(0.08206 L*atm/mol*k)(277.15) 1.57×106 L = 22.742929(x) x=69032.44521 mol O2 69032.44521 mol O2 * 2 mol KClO3 / 3 mol O2 = 138064.8904 mol KClO3 138064.8904 mol KClO3 * 122.548 g / 1 mol KClO3 16919576.19 g KClO3 Trying to make sure my calculations are right. Thanks

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Question

If an inflatable device has a volume of 1.57 x 106 L. And in order to use enough O2 to

inflate it we use this reaction:

2 KClO3 (s) → 3 O2 (g) + 2 KCl (s)

If the temperature is 4.00 C and the pressure is 1.00 atm, how many grams of KClO3 (s) are required to produce 1.57 x 106 L of O2 (g)? R=0.08206 Lxatm/molxk

Pv=nrt

(1.00atm)(1.57×106 L)=(x)(0.08206 L*atm/mol*k)(277.15)

1.57×106 L = 22.742929(x)

x=69032.44521 mol O2

69032.44521 mol O2 * 2 mol KClO3 / 3 mol O2 = 138064.8904 mol KClO3

138064.8904 mol KClO3 * 122.548 g / 1 mol KClO3

16919576.19 g KClO3

Trying to make sure my calculations are right. Thanks

 
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