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Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 13.3 mL of water (d= 1.00 g/mL) and 24.1 g of SO3?
How much of the reagent in excess is left over?

Respuesta :

SkyAce

The answer is 31.5g. See solution below

SO3 + H2O => H2SO4

Theoretical moles of SO3 : H2O = 1 : 1

Moles of SO3 = mass/molar mass of SO3

= 25.7/80.064 = 0.3210 mol

Mass of H2O = volume x density

= 13.9 x 1.00 = 13.9 g

Moles of H2O = mass/molar mass of H2O

= 13.9/18.02 = 0.7714 mol

Experimentall moles of SO3 : H2O = 0.3210 : 0.7714

= 0.416 : 1 = 1 : 2.40

Since H2O is in excess, SO3 is the limiting reactant

Moles of H2SO4 = Moles of SO3 = 0.3210 mol

Mass of H2SO4 = moles x molar mass of H2SO4

= 0.3210 x 98.08

= 31.5 g

I hope this helps!

pstnonsonjoku

0.44 moles or 7.92 g of excess reagent is left over.

We must first write down the equation of the reaction;

SO3(g) + H2O(l) → H2SO4(aq)

We can obtain the mass of water from its density as follows;

mass = density × volume

mass = 1.00 g/mL ×   13.3 mL = 13.3 g of water

number of moles of water = mass/molar mass = 13.3 g/18 g/mol = 0.74 moles

Number of moles of SO3 = 24.1 g/80 g/mol = 0.30 moles

To find the limiting reactant, we can see that the reaction is 1:1 so we need 0.30 moles of SO3 to react with 0.30 moles of water but water is clearly in excess so SO3 is the limiting reagent.

Amount of excess reagent left over = 0.74 moles - 0.30 moles = 0.44 moles of excess reagent left over.

Mass left over = 0.44 moles ×  18 g/mol = 7.92 g

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