Question

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

Answer 1

Answer:

13.4g

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

$x = \frac{7.1 \times 10²² }{ 6.02 × 10²³}$

$x = \frac{71}{602}$

we now have the number of moles of Al₂CO₃

to calculate the grams (mass):

$moles = \frac{mass}{relative \: formula \: mass}$

$mass = moles \: \times \: relative \: formula \: mass$

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:

$\frac{71}{602} \times 114 = 13.4g$