Question

How many grams of Al2O3are present in 7.1 x 10^22 molecules of Al2O3?

Answer 1

13.4g

Step-by-step explanation:

we know that:

1 mole = 6.02 × 10²³ atoms

make the unknown number of moles = x

x = 7.1 × 10²² atoms

putting them both together:

1 mole = 6.02 × 10²³ atoms

x = 7.1 × 10²² atoms

Cross multiply:

6.02 × 10²³ x = 7.1 × 10²²

divide both sides by 6.02 × 10²³

`x = (7.1 * 10²² )/( 6.02 × 10²³)`

`x = (71)/(602)`

we now have the number of moles of Al₂CO₃

to calculate the grams (mass):

`moles = (mass)/(relative \: formula \: mass)`

`mass = moles \: * \: relative \: formula \: mass`

add up all of the atomic masses of Al₂CO₃ to calculate relative formula mass:

(27 × 2) + 12 + (16 × 3) = 114

the grams (mass) of Al₂CO₃:

`(71)/(602) * 114 = 13.4g`