Question

Be sure to answer all parts. Butane gas is compressed and used as a liquid fuel in disposable cigarette lighters and lightweight camping stoves. Suppose a lighter contains 5.10 mL of butane (d-0.579 g/mL). (a) How many grams of oxygen are needed to burn the butane completely? g 02 (b) How many moles of H2O form when all the butane burns?" moles H20 (c) How many total molecules of gas form when the butane burns completely? x 10 (rclect) molecules of gas (Enter your answer in scientific notation.)

Answer 1

volume of butane = 5.10ml

density of butane   = 0.579g/ml

mass of butane = volume * desnity

                         = 5.10*0.579   = 2.9529g

a.

2C4H10(g) + 26O2(g) -------------------> 8CO2(g) + 10H2O(g)

2 moles of C4H10 react with 26 moles of O2

2*58g of C4H10 react with 26*32g of O2

2.9529g of C4H10 react with = 26*32*2.9529/(2*58)   = 21.18g of O2

b.

2C4H10(g) + 26O2(g) -------------------> 8CO2(g) + 10H2O(g)

2 moles of C4H10 react oxygen to gives 10 moles of H2O

2*58g of C4H10 react with oxygen to gives 10 moles of H2O

2.9529g of C4H10 react with oxygen to gives = 10*2.9529g/(2*58g)   = 0.2545 moles of H2O

c.

2C4H10(g) + 26O2(g) -------------------> 8CO2(g) + 10H2O(g)

2 moles of C4H10 react oxygen to gives 8 moles of CO2

2*58g of C4H10 react with oxygen to gives 8 moles of CO2

2.9529g of C4H10 react with oxygen to gives   = 8*2.9529/(2*58)   = 0.2036moles of CO2

total no of moles = 0.2545 mole + 0.2036 moles   = 0.4581 moles

total no of molecules = no of moles * 6.023*10^23

                                 = 0.4581*6.023*10^23

                                 = 2.76*10^23 molecules >>>>>answer