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Pani-rosa [81]
3 years ago
14

Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. Based on the r

eactant used in your experiment, do you think it is reasonable to exclude these types of substances as contaminants that would give a false positive when you tested your reaction residue to verify that it is KCl?
Chemistry
1 answer:
SashulF [63]3 years ago
3 0

Answer:

Other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide. However iodides and bromides have different colours hence they will not give a false positive test for KCl. Other chlorides present may lead to a false positive test for KCl.

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution is used. Various halide ions give various colours of precipitate with silver nitrate. Chlorides yield a white precipitate, bromides yield a cream precipitate while iodides yield a yellow precipitate. All these ions or some of them may be present in the system.

However, if other chlorides are present, they will also yield a white precipitate just as KCl leading to a false positive test for KCl. Since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. We can exclude other halides from the tendency to lead us to a false positive test for KCl but not other chlorides.

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3 years ago
What is the mass of the solid NH4Cl formed when 75.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas
Gekata [30.6K]

Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of NH_4Cl produced is, 110.7 grams.

Explanation :

The balanced chemical reaction will be:

NH_3+HCl\rightarrow NH_4Cl

First we have to calculate the moles of NH_3 and HCl

\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}

Molar mass of NH_3 = 17 g/mole

\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole

and,

\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}

Molar mass of HCl = 36.5 g/mole

\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NH_3

So, 2.07 mole of HCl react with 2.07 mole of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of gas = 752 mmHg = 0.989 atm     (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 14.0^oC=273+14.0=287K

Putting values in above equation, we get:

0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K

V = 56.5 L

Now we have to calculate the moles of NH_4Cl

As, 1 mole of HCl react with 1 mole of NH_4Cl

So, 2.07 mole of HCl react with 2.07 mole of NH_4Cl

Now we have to calculate the mass of NH_4Cl

\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl\times \text{ Molar mass of }NH_4Cl

Molar mass of NH_4Cl = 53.5 g/mole

\text{ Mass of }NH_4Cl=(2.07moles)\times (53.5g/mole)=110.7g

Thus, the volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of NH_4Cl produced is, 110.7 grams.

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3 years ago
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