Sulphuric acid decomposes into H2O and SO3. H2SO4(l)⟶H2O(g)+SO3(g) If 10.0 g of sulphuric acid is completely decomposed in a closed container with a total volume of 3.00 L and no air at 350∘C, what will the total pressure in the container be in atmospheres? Assume the volume of the liquid is negligible. Answer should have three significant figures.

3.48 atm

Explanation:

The first step is to determine the number moles of sulfuric acid. To do this find the molar mass of sulfuric acid.

Molar Mass of H2SO4 = (1 g/mol*2) + (32 g/mol) + (16 g/mol * 4) = 98 g/mol

Then divide the mass of sulfuric acid by the molar mass of the sulfuric acid to get the number of moles of sulfuric acid.

(10 g)/(98 g/mol) = 0.10204 mol

Then use the molar ratios to determine the moles of gas produced.  In this case all of the ratios are 1 to 1 meaning that 0.10204 mol of H2SO4 produces 0.10204 mol of H2O and 0.10204 mol of SO3.  That means that 0.2041 moles of gas are produced as a result of the reaction.

The next thing to do is to convert the temperature to Kelvin since gas equations need Kelvin.  To do this add 273.15.

350 degrees C + 273 = 623.15 K

Now the PV=nRT can be used and when solving for pressure the equation can be written P=nRT/V.  

P = pressure and is the unknown

V = volume and is 3.00 L

n = moles of gas and is 0.2041 mol

R = gas constant and is 0.08205 L atm/mol K

T = temperature and is 623.15 K

P = (0.2041 mol * 0.08205 L atm/mol K * 623.15 K) / 3.00 L

P = 3.4785 atm which is 3.48 atm when rounded to three significant figures

I hope this helps.  Let me know if anything is unclear.

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