Question

use the references to access important values if needed for this question.
how many grams of ag₂co₃ will precipitate when excess k₂co₃ solution is added to 66.0 ml of 0.748 m agno₃ solution?
2agno₃(aq) + k₂co₃(aq) → ag₂co₃(s) + 2kno₃(aq)

Explanation:

Step1: Calculate the moles of \(AgNO_3\)

Use the formula \(n = C\times V\). Given \(C = 0.748\space M\) and \(V=66.0\space mL = 66.0\times10^{- 3}\space L\).
\(n(AgNO_3)=0.748\space M\times66.0\times 10^{-3}\space L = 0.049368\space mol\)

Step2: Determine the mole ratio

From the balanced chemical equation \(2AgNO_3(aq)+K_2CO_3(aq)\to Ag_2CO_3(s)+2KNO_3(aq)\), the mole ratio of \(AgNO_3\) to \(Ag_2CO_3\) is \(2:1\). So \(n(Ag_2CO_3)=\frac{1}{2}n(AgNO_3)\)
\(n(Ag_2CO_3)=\frac{1}{2}\times0.049368\space mol = 0.024684\space mol\)

Step3: Calculate the mass of \(Ag_2CO_3\)

The molar mass of \(Ag_2CO_3\) is \(M=(2\times107.87 + 12.01+3\times16.00)\space g/mol=(215.74 + 12.01+48.00)\space g/mol = 275.75\space g/mol\)
Use the formula \(m = n\times M\), so \(m(Ag_2CO_3)=0.024684\space mol\times275.75\space g/mol\approx6.81\space g\)

Answer:

\(6.81\space g\)