KCET · Chemistry · Some Basic Concepts of Chemistry
The mass of \( \mathrm{AgCl} \) precipitated when a solution containing \( 11.70 \mathrm{~g} \) of \( \mathrm{NaCl} \) is added to a
solution containing \( 3.4 \mathrm{~g} \) of \( \mathrm{AgNO}_{3} \) is [Atomic mass of \( \mathrm{Ag}=108 \), Atomic mass of \( \mathrm{Na}=23 \) ]
- A \( 2.87 \mathrm{~g} \)
- B \( 6.8 \mathrm{~g} \)
- C \( 5.74 \mathrm{~g} \)
- D \( 1.17 \mathrm{~g} \)
Answer & Solution
Correct Answer
(A) \( 2.87 \mathrm{~g} \)
Step-by-step Solution
Detailed explanation
Balanced reactio is:
\(\mathrm{NaCl}+\mathrm{AgNO}_{3} \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_{3}\)
molar mass of \(\mathrm{AgNO}_{3}=170 \mathrm{~g}\)
molar mass of \(\mathrm{NaCl}=58.5 \mathrm{~g}\)
molar mass of \(\mathrm{AgCl}=143.5 \mathrm{~g}\)
from the reaction \(58.5 \mathrm{~g} \mathrm{NaCl}\) reacts with \(170 \mathrm{~g} \mathrm{AgNO}_{3}\) thus \(5.85 \mathrm{~g} \mathrm{NaCl}\) will reacts with \(=\frac{170}{58.5} \times 5.85=17 \mathrm{~g} \mathrm{AgNO}_{3}\)
Given \(\mathrm{AgNO}_{3}=3.4 \mathrm{~g}\)
therefore \(\mathrm{AgNO}_{3}\) is completely consumed and is the limiting reagent
\(170 \mathrm{~g} \mathrm{AgNO}{ }_{3}\) produces \(143.5 \mathrm{~g} \mathrm{AgCl}\)
\(3.4 \mathrm{~g}\) will produce \(=\frac{143.5}{170} \times 3.4=2.87 \mathrm{~g} \mathrm{} \mathrm{AgCl}\)
Hence, option A is correct.
\(\mathrm{NaCl}+\mathrm{AgNO}_{3} \rightarrow \mathrm{AgCl}+\mathrm{NaNO}_{3}\)
molar mass of \(\mathrm{AgNO}_{3}=170 \mathrm{~g}\)
molar mass of \(\mathrm{NaCl}=58.5 \mathrm{~g}\)
molar mass of \(\mathrm{AgCl}=143.5 \mathrm{~g}\)
from the reaction \(58.5 \mathrm{~g} \mathrm{NaCl}\) reacts with \(170 \mathrm{~g} \mathrm{AgNO}_{3}\) thus \(5.85 \mathrm{~g} \mathrm{NaCl}\) will reacts with \(=\frac{170}{58.5} \times 5.85=17 \mathrm{~g} \mathrm{AgNO}_{3}\)
Given \(\mathrm{AgNO}_{3}=3.4 \mathrm{~g}\)
therefore \(\mathrm{AgNO}_{3}\) is completely consumed and is the limiting reagent
\(170 \mathrm{~g} \mathrm{AgNO}{ }_{3}\) produces \(143.5 \mathrm{~g} \mathrm{AgCl}\)
\(3.4 \mathrm{~g}\) will produce \(=\frac{143.5}{170} \times 3.4=2.87 \mathrm{~g} \mathrm{} \mathrm{AgCl}\)
Hence, option A is correct.
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