If 50.0 dm3 of methane, ch4, react with 10.0 dm3 of air, calculate the grams of water produced.
ch4 (g) + 2 o2 (g) --> co2 (g) + 2 h2o (l)

Answer : The mass of produced will be, 8.1 grams.

Explanation : Given,

Volume of =    

Volume of =

Molar mass of = 18 g/mole

First we have to calculate the moles of and .

As, 22.4 L volume of present in 1 mole of

So, 50 L volume of present in mole of

and,

As, 22.4 L volume of present in 1 mole of

So, 10 L volume of present in mole of

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 moles of react with 1 mole of

So, 0.45 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of .

As, 2 moles of react to give 2 moles of

As, 0.45 moles of react to give 0.45 moles of

Now we have to calculate the mass of .

Therefore, the mass of produced will be, 8.1 grams.

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

pie chart because it is in percents ( % )