Chemistry, 03.07.2019 19:20 ernestinebell3514

An impure sample of table salt that weighed 0.8421 g, when dissolved in water and treated with excess agno3, formed 2.044 g of agcl. what is the percentage of nacl in the impure sample?

Answers: 3

Show answers

Answers

Answer from: 5001Je

An impure sample of table salt that weighed 0.8421 g and treated with excess AgNO₃ formed 2.044 g of AgCl, has a percentage of NaCl of 98.98%.

Let's consider the reaction between NaCl and AgNO₃ to produce AgCl and NaNO₃.

NaCl + AgNO₃ ⇒ AgCl + NaNO₃

We can calculate the mass of NaCl that produced 2.044 g of AgCl using the following relations.

The molar mass of AgCl is 143.32 g/mol.The molar ratio of AgCl to NaCl is 1:1.The molar mass of NaCl is 58.44 g/mol.

$2.044 g AgCl \times \frac{1molAgCl}{143.32 g AgCl} \times \frac{1molNaCl}{1molAgCl} \times \frac{58.44gNaCl}{1molNaCl} = 0.8335gNaCl$

An impure sample of mass 0.8421 g contains 0.8335 g of NaCl. The percentage of NaCl in the impure sample is:

$\% NaCl = \frac{0.8335g}{0.8421g} \times 100\% = 98.98\%$

An impure sample of table salt that weighed 0.8421 g and treated with excess AgNO₃ formed 2.044 g of AgCl, has a percentage of NaCl of 98.98%.

Learn more: link

Answer from: gl648809

99.24%.

Explanation:

NaCl reacted with AgNO₃ as in the balanced equation:

NaCl + AgNO₃ → AgCl(↓) + NaNO₃,

1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.

We need to calculate the no. of moles of AgCl produced:

no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.

Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:

using cross multiplication:

1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.

∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.

Now, we can get the mass of puree NaCl in the sample:

mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.

∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.

Answer from: Quest

the answer is learned

Answer from: Quest

explanation:

Another question on Chemistry

Chemistry, 22.06.2019 14:30

Chemistry worksheet - i am not sure what they are asking for exactly?

Answers: 1

Answer

Chemistry, 22.06.2019 16:30

How do i turn 2 s into one in the products

Answers: 1

Answer

Chemistry, 22.06.2019 19:10

How does the atmosphere to make earth livable? check all that apply. causes the seasons contains oxygen provides warmth creates important nutrients blocks harmful energy from the sun plz like !

Answers: 2

Answer

Chemistry, 23.06.2019 02:30

Calculate the ph at the equivalence point for the titration of a solution containing 150.0 mg of ethylamine (c2h5nh2) with 0.1000 m hcl solution. the volume of the solution at the equivalence point is 250.0 ml. kb forethylamine is 4.7 × 10−4 .

Answers: 2

Answer

You know the right answer?

An impure sample of table salt that weighed 0.8421 g, when dissolved in water and treated with exces...

Questions