When 5 g of barium chloride reacted with 3 g of sodium sulfate the products formed were 3.5 g of barium sulfate and x gram of sodium chloride what will be the amount of sodium chloride ph5 g of barium chloride reacted with 3 g of sodium sulfate the products formed were 3.5 g of barium sulfate and x gram of sodium chloride what will be the amount of sodium chloride formed

The amount of sodium chloride formed is 0.03 moles which are contained in 1.75 g

Explanation:

The reaction is this:

BaCl₂  + Na₂SO₄ → BaSO₄ + 2NaCl

5g             3g            3.5g          x

First of all, let's calculate the moles of everything

Mol = Mass / Molar mass

5g / 208.23 g/m = 0.024 mol  BaCl₂

3g / 142.06 g/m = 0.0211 mol Na₂SO₄

3.5g / 233.39 g/m = 0.0150 mol BaSO₄

We can now determinate, the limiting reactant to work

Ratio between reactants is 1:1, so 1 mol of chloride reacts with 1 mol of sulfate. Limiting agent is the Na₂SO₄ ( I need 0.024 mol to react the BaCl₂, and I only have 0.0211 mol, BaCl₂ will remain without reaction)

Ratio is 1:1 between the sulfates so, 0.0211 mol of sodium sulfate produce 0.0211 mol of barium sulfate but I only produced 0.0150 mole of it.

This reaction has a % of yield.

0.0211 moles ___ 100 %

0.0150 moles ___ (0.0150 . 100)/0.0211 = 71.09 %

Now we can know the production of NaCl

Ratio is 1:2, so If i have 0.0211 mol of sodium sulfate I'll produce the double of NaCl (0.0211 .2) = 0.0422 mol

As the reaction has a 71.09% yield, I'll produce

0.0422 mol . 0.7109 = 0.03mol

Molar mass NaCl = 58.45 g/m

0.03 m . 58.45 g/m = 1.75 g

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