Butane gas is compressed and used as liquid fuel in cigarrettelighters. Suppose
ID: 688602 • Letter: B
Question
Butane gas is compressed and used as liquid fuel in cigarrettelighters. Suppose a lighter contains 5.75 mL (d=0.579 g/mL)A) How many grams of oxygen are needed to burn the butanecompletely?
B) How many moles of H2Oform when the butane is burned completely?
C) How many total molecules of gas form when all the butaneburns?
A) How many grams of oxygen are needed to burn the butanecompletely?
B) How many moles of H2Oform when the butane is burned completely?
C) How many total molecules of gas form when all the butaneburns?
Explanation / Answer
A) Volume ofa fuel in lighter V = 5.75 mL density offuel d = 0.579g/mL we know therelation d= mass / volume massof fuel m = d*volume =0.579gmL*5.75mL = 3.33g Therefore mass of butane gas m= 3.33g The reaction between butane andO2is 2C4H10(g) + 13O2(g)--------> 8CO2(g) +10H2O(l)so 2*58g of butane is needed to 13*32gof Oxygen for completed the reaction. now the required oxygen for complete the reaction with3.33 g of butane is = ( 13*32gof O2 / 2*58g of C4H10)*3.33g ofC4H10 = 11.94 g of O2 The required oxygen is 11.94g to complethe reaction . B) From the above equation you know that
10mol of H2Oformed by 2.0mol of butane reacted. So from 3.33g of butane mole ofH2O formed is 3.33g ofbutane in mol =3.33g /58g/mol = 0.0574mol of butane molH2O formed =(10mol of H2O /2.0mol ofbutane)* 0.0574mol butane = 0.287mol of H2O moles of H2O formed in thereaction = 0.287mol C) If all the butane gas was burnt ,(0.37*6.023*1023 ) molecules of gas formed.
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