calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

21.2 gm

Explanation:

calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units

butane is the hydrocarbon C4H10  

in combustion, we react hydrocarbons with O2 to form CO2 and H2O

so

C4H10  + O2>  CO2 + H2O

BALANCE

2C4H10 + 1302> 8CO2 + 10 H2O

the molar mass of CO2 is 12 + 16X2 = 44

64.1 gm of CO2 is

64.1/44 = 1.46 MOLES OF  CO2,

FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE  IT IS A

8:2 OR 4:1 RATIO.  THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2

SO

THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES

THE MOLAR MASS OF BUTANE IS 58.12

0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm