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?

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!

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years