Question

For the reaction below, if 6.3 g of S reacted with 10.0 g of O₂, how many grams of SO3 will be produced?

2S (s) + 30₂(g) → 2S03 (g)

Answer 1

2S + 302 = 2SO3

Mass of S = 6.3g

Mass of 02 = 10.0g

n = m/M

M(S) = 32g/mol

n = 6.3g/32g/mol

n = 0.195mol

n(S)/n(SO3) = 2/2

Let n(SO3) = x

2(0.195) = 2x

0.39 = 2x

x = 0.195

Therefore, n(SO3) = 0.195mol

For mass of SO3

m = M×n

But M(SO3) = (32×1) + (16×3)

= 80g/mol

m = 80g/mol × 0.195mol

m = 15.6g

Therefore, 15.6g of SO3 will be produced.

HOPE IT HELPS. HAVE A WONDERFUL DAY.