Answer:
44 g CO₂
Explanation:
The chemical equation for the reaction is:
C(s) + O₂(g) → CO₂(g)
Thus, 1 mol of C(s) reacts with 1 mol of O₂. We convert the moles to grams by using the molecular weight of each compound (C= 12 g/mol; O₂= 32 g/mol):
1 mol C(s) = 12 g/mol x 1 mol = 12 g
1 mol O₂= 32 g/mol x 1 mol = 32 g
The stoichiometric ratio C(s)/O₂ is 12 g C/32 g O₂. We have 40 g of O₂, so it will be consumed:
40 g O₂ x 12 g C/32 g O₂= 15 g C(s)
And we have 12 g. Thus, the limiting reactant is C(s).
Now, we use the limiting reactant to calculate the mass of product produced (CO₂)
Molecular weight CO₂ = 12 g/mol x 1 C + (16 g/mol x 2 O) = 44 g/mol
1 mol CO₂ = 44 g/mol x 1 mol = 44 g
So, 44 grams of CO₂ are produced from 12 g of C(s) (44 g CO₂/12 g C). So, the amount produced when 12 g of C(s) reacts with 40 g O₂ is:
12 g C(s) x 44 g CO₂/12 g C(s) = 44 g
