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.

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).

po-215 i got it right on the quiz