DP Chemistry (last assessment 2024)

Question 21M.2.hl.TZ1.4c(ii)

Date	May 2021	Marks available	[Maximum mark: 3]	Reference code	21M.2.hl.TZ1.4c(ii)
Level	hl	Paper	2	Time zone	TZ1
Command term	Calculate	Question number	c(ii)	Adapted from	N/A

c(ii).

[Maximum mark: 3]

21M.2.hl.TZ1.4c(ii)

Hydrogen peroxide can react with methane and oxygen to form methanol. This reaction can occur below 50°C if a gold nanoparticle catalyst is used.

Methanol is usually manufactured from methane in a two-stage process.

CH₄(g) + H₂O (g) $⇌$ CO (g) + 3H₂(g)
CO (g) + 2H₂(g) $⇌$ CH₃OH (l)

(c(ii))

8.00 g of methane is completely converted to methanol. Calculate, to three significant figures, the final volume of hydrogen at STP, in dm³. Use sections 2 and 6 of the data booklet.

[3]

Markscheme

amount of methane = « $\frac{8.00 g}{16.05 g {mol}^{- 1}}$ = » 0.498 «mol» ✔

amount of hydrogen = amount of methane / 0.498 «mol» ✔

volume of hydrogen = «0.498 mol × 22.7 dm³mol⁻¹ = » 11.3 «dm³» ✔

Award [3] for final correct answer.
Award [2 max] for 11.4 «dm³ due to rounding of mass to 16/moles to 0.5. »

Syllabus sections

Core

Core » Topic 1: Stoichiometric relationships » 1.3 Reacting masses and volumes

Core » Topic 1: Stoichiometric relationships

Question 21M.2.hl.TZ1.4c(ii)

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