ARCH-C-04: Molar Gas Volume (24 dm^3 at rtp), Ideal Gas Scaling, Percentage Yield & Atom Economy

Full 15-Q Timed Drill

Step 1 of 6 • The Core Idea

ARCH-C-04: Why the 150s Classroom Method Fails Under the 90s Clock

Standard Classroom Method (Too Slow)~150s (Blows 90s Clock)

First calculate the moles of magnesium:

n(\text{Mg}) = \frac{0.60}{24} = 0.025\text{ mol}

. From the

1:1

stoichiometry,

n(\text{H}_2) = 0.025\text{ mol}

. At rtp,

V(\text{H}_2) = 0.025 \times 24\text{ dm}^3 = 0.60\text{ dm}^3 = 600\text{ cm}^3

(~105s).

The Fast ShortcutTarget: 40s (Saves 110s)

**Line 1 (

24

Cancellation):**

V(\text{cm}^3) = \frac{m}{A_r} \times 24,000 = \frac{0.60}{24} \times 24,000

.
**Line 2 (Cancel

24

):**

0.60 \times 1000 = 600\text{ cm}^3

(20s).

Where Students Lose Time: In 18 past UAT-UK questions (8% of Chemistry), candidates lose 110s by mixing cm³ and dm³ without applying 1 dm³ = 1000 cm³. Instead, apply C4 with Maths 1 skill M1.2.

Step 1 of 6 • Press ArrowRight or Space