If you burned 4.10 x 10^24 molecules of butane (C4H10), what mass of ethane did you burn?

Explanation:

Molar mass of C4H10 = 12x 4 + 1 x 10= 48+ 10=58g.

Also note that molar mass of a substance is equivalent to Avogadro's number (6.02 X 10^-23)

Hence,

6.02 X 10^-23 molecules of ethane was burned in 58g of ethane

So, 4.10 X 10^24 molecules of ethane will burn in xg of ethane.

Cross multiply:

6.02 X 10^-23x = 58 X 4.10 X 10^24

6.02 X 10^-23x = 2.378 X 10 ^26

Dividing both sides by the coefficient of 'x'

x = 2.378 X 10^26/(6.02X 10^-23)

x = 3.95 X10^48g approximately

x = 4.0 X 10^48g

Or

Mole = number of elementary

particless / Avogadro's number

Where,

No of elementary particles= 4.10 X 10^24 molecules

Avogadro's number= 6.02 x 10^-23

Therefore,

Mole = 4.10 X 10^24/ (6.02 X 10^-23)

= 6.81 X 10^ 46 moles.

But mole = mass / molar mass

Mass = mole x molar mass

where

Molar mass of C4H10 = 12x 4 + 1 x 10= 48+ 10=58g/mol

Mass= 6.81 X 10^46 X 58

= 3.95 X 10^48g approximately

= 4.0 X 10^ 48g.

Thanks

the answer is d. 74.02