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
[tex]x = \frac{7.1 \times 10²² }{ 6.02 × 10²³} [/tex]
[tex]x = \frac{71}{602} [/tex]
we now have the number of moles of Al₂CO₃
to calculate the grams (mass):
[tex]moles = \frac{mass}{relative \: formula \: mass} [/tex]
[tex]mass = moles \: \times \: relative \: formula \: mass[/tex]
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:
[tex] \frac{71}{602} \times 114 = 13.4g[/tex]
