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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

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