An equimolar mixture of NaCl and KCl weigh 266mg .Find the mass of AgCl produced on adding excess AgNO3 to above mixture?

Question

Answer

Explanation:

Step 1: Calculate the moles of NaCl and KCl:

Given, the mixture contains NaCl and KCl in equimolar amounts. Let the number of moles of NaCl and KCl be 'x'. Therefore, the total number of moles in the mixture is 2x.

Molar mass of NaCl = 58.44 g/mol

Molar mass of KCl = 74.55 g/mol

The weight of the mixture is 266 mg i.e. 0.266 g. Therefore, the number of moles of the mixture is:

n = m/M = 0.266/(58.44 + 74.55) = 0.00176 moles

Since the mixture contains equal amounts of NaCl and KCl, each compound must have 0.00088 moles.

Step 2: Write the balanced chemical equation for the reaction between AgNO3 and NaCl/KCl:

AgNO3 + NaCl → AgCl + NaNO3
AgNO3 + KCl → AgCl + KNO3

Step 3: Calculate the moles of AgNO3 required to react with NaCl/KCl:

Since the mixture contains 0.00088 moles of NaCl and KCl, we need to add an excess amount of AgNO3 to react with both the compounds completely.

Molar mass of AgNO3 = 169.87 g/mol

The number of moles of AgNO3 required is:

n = 2 × 0.00088 = 0.00176 moles

The mass of AgNO3 required is:

m = n × M = 0.00176 × 169.87 = 0.299 g

Step 4: Calculate the mass of AgCl produced:

Since AgNO3 is in excess, all the NaCl and KCl will react with AgNO3 to form AgCl.

Molar mass of AgCl = 143.32 g/mol

The number of moles of AgCl produced is:

n = 0.00088 × 2 = 0.00176 moles

The mass of AgCl produced is:

m = n × M = 0.00176 × 143.32 = 0.253 g

Therefore, the mass of AgCl produced on adding excess AgNO3 to the given mixture is 0.253 g.

Answer

574mg

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