Bobby hill bought a can of butane (C4H10) for his portable stove, and each can c
ID: 635387 • Letter: B
Question
Bobby hill bought a can of butane (C4H10) for his portable stove, and each can contains 250g of butane and has a volume of 250ml.h aoaca of butane(CHn) for his portable stove, and each can contains 250g of butane (Spts) What is the pressure inside the can when it is stored at room temperature(298K)? a. 250 x o.oszlatn zask 25O V2sog Il o.oszi T-299K (Spts) Bobby burned cooking? b. 100g of butane to cook some dinner, what is the new pressure inside the can after h at R-o.oe (10pts) How many grams of oxygen would be required to burn the 100g of butane that Bobby used? c. 0 d. (5pts) KCIO, decomposes into KCl and Os, write a balanced equation for the reaction. (10pts) If 257.8g of KCIO, was used as the only source of oxygen, would there be enough oxygen produced to burn the 100g of butane that Bobby used? Explain your answer. e. fe bvrm the oet ben usec 2 TOTAL: /100
Explanation / Answer
Ans 2
Part a
Moles of butane = mass/molecular weight
= 250g / 58.12 g/mol
= 4.3014 mol
Volume V = 250 mL x 1L/1000 mL = 0.550 L
From the ideal gas equation
P = nRT/V
= 4.3014 mol x 0.0821 L-atm/mol-K x 298K / 0.250 L
= 420.95 atm
Part b
Butane burned = 100 g
Butane remain = 250 - 100 = 150 g
Moles of butane = 150g / 58.12 g/mol
= 2.581 mol
Volume V = 250 mL x 1L/1000 mL = 0.550 L
From the ideal gas equation
P = nRT/V
= 2.581 mol x 0.0821 L-atm/mol-K x 298K / 0.250 L
= 252.58 atm
Part c
The balanced reaction with molar mass
2 C4H10 + 13 O2 = 8 CO2 + 10 H2O
MM: 58.12 32 44 18
(2*58.12=) 116.24 g butane required = (32*13=) 416 g O2
100 g butane required = 416*100/116.24 = 357.88 g O2
Part d
The balanced decomposition reaction with molar mass
2 KClO3 = 2 KCl + 3 O2
MM : 122.55 74.55 32
(2*122.55=) 245.1 g KClO3 produces = (32*3=) 96 g O2
257.8 g KClO3 produces = 96*257.8/245.1 = 100.97 g O2
We require O2 for 100 g butane combustion = 357.88 g O2
Therefore 257.8 g KClO3 is not enough.