Question

a student determines the silver content of a solution by first precipitating it as silver hydroxide, and then decomposing the hydroxide to silver oxide by heating. how many grams of silver oxide should the student obtain if his solution contains 56.0 ml of 0.583 m silver nitrate?
use the references to access important values if needed for this question.

Explanation:

Step1: Calculate the moles of silver nitrate

The formula for moles \(n = c\times V\). Given \(c = 0.583\space M\) and \(V=56.0\space mL=0.056\space L\).
\(n = 0.583\times0.056\)
\(n = 0.032648\space mol\)

Step2: Write the chemical equations

The precipitation reaction: \(AgNO_{3}+NaOH = AgOH\downarrow+NaNO_{3}\)
The decomposition reaction: \(2AgOH=Ag_{2}O + H_{2}O\)
From the equations, \(2\space mol\space AgNO_{3}\) gives \(1\space mol\space Ag_{2}O\)

Step3: Calculate the moles of silver oxide

If \(n(AgNO_{3}) = 0.032648\space mol\), then \(n(Ag_{2}O)=\frac{0.032648}{2}= 0.016324\space mol\)

Step4: Calculate the mass of silver oxide

The molar mass of \(Ag_{2}O\) is \(M=(2\times107.87 + 16)=231.74\space g/mol\)
The mass \(m=n\times M\)
\(m = 0.016324\times231.74\)
\(m\approx3.78\space g\)

Answer:

\(3.78\space g\)