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aev [14]
3 years ago
15

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

Chemistry
1 answer:
Alexus [3.1K]3 years ago
5 0

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

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