Brainliest for an answer! 2 As (s) + 6 NaOH (aq) → 2 Na3AsO3 (s) + 3 H2 (g)

If 34.0g of As react with excess NaOH at STP, what volume of H2 will be produced in liters?

If you can answer this question quickly, I will mark you brainliest, I'm stumped to be honest.

The volume of H₂ : = 15.2208 L

Given

Reaction

2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)

34.0g of As

Required

The volume of H₂ at STP

Solution

mol As (Ar = 75 g/mol) :

= mass : Ar

= 34 g : 75 g/mol

= 0.453 mol

From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :

=3/2 x mol As

=3/2 x 0.453

= 0.6795

At STP, 1 mol = 22.4 L, so :

= 0.6795 x 22.4 L

= 15.2208 L

the answer is learned