Answer:
d
Step-by-step explanation:
2H₂(g) +CO(g) → CH₄O (l)
Since H₂ was reacted with excess CO, H₂ is the limiting reagent. This means that the amount of product formed is dependent on the amount, i.e. the number of moles, of H₂.
The number of moles of H₂ can be calculated by taking the mass divided by its Mr (relative molecular mass).
Mr of H₂= 2(1)= 2
Moles of H₂ (g)
= 6.12 ÷2
= 3.06 mol
From the balanced equation, 2 moles of H₂(g) produces 1 mole of CH₄O(g).
Since we have 3.06 moles of H₂(g),
Moles of CH₄O (l) produced
= 3.06 ÷2
= 1.53 mol
Mr of CH₄O
= 12 +4 +16
= 32
Mass= mole ×Mr
Mass of CH₄O (l) produced
= 1.53(32)
= 48.96 g
Thus, d would be the best option.