Question

use the references to access important values if needed for this question. according to the following reaction, how many grams of iron(ll) nitrate will be formed upon the complete reaction of 27.6 grams of silver nitrate with excess iron(ll) chloride? iron(ll) chloride (aq) + silver nitrate (aq) → iron(ll) nitrate (aq) + silver chloride (s) mass = g submit answer retry entire group 2 more group attempts remaining the equation for this reaction is: fecl₂(aq) + 2agno₃(aq) → fe(no₃

Explanation:

Step1: Calculate moles of \(AgNO_3\)

The molar mass of \(AgNO_3\) is \(M = 107.87+14 + 3\times16=169.87\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 27.6\space g\) and \(M = 169.87\space g/mol\), we have \(n_{AgNO_3}=\frac{27.6}{169.87}\space mol\).

Step2: Use mole - ratio from balanced equation

From the balanced equation \(FeCl_2(aq)+2AgNO_3(aq)\to Fe(NO_3)_2(aq) + 2AgCl(s)\), the mole - ratio of \(AgNO_3\) to \(Fe(NO_3)_2\) is \(2:1\). So \(n_{Fe(NO_3)_2}=\frac{1}{2}n_{AgNO_3}=\frac{1}{2}\times\frac{27.6}{169.87}\space mol\)

Step3: Calculate mass of \(Fe(NO_3)_2\)

The molar mass of \(Fe(NO_3)_2\) is \(M = 55.85+2\times(14 + 3\times16)=179.85\space g/mol\).
Using the formula \(m = n\times M\), we substitute \(n=\frac{1}{2}\times\frac{27.6}{169.87}\space mol\) and \(M = 179.85\space g/mol\)

$$
LATEXBLOCK0
$$

Answer:

\(14.6\space g\)