...agno₃(aq) +...bacl₂(aq) →... how many grams of barium chloride is necessary to react wit 5.0 g of silver…

Explanation:

Step1: Balance the chemical equation

The reaction between $\ce{AgNO3}$ and $\ce{BaCl2}$ is a double - displacement reaction. The balanced equation is: $$2\ce{AgNO3(aq)}+\ce{BaCl2(aq)}\rightarrow 2\ce{AgCl(s)}+\ce{Ba(NO3)2(aq)}$$

Step2: Calculate moles of $\ce{AgNO3}$

The molar mass of $\ce{AgNO3}$ is $M(\ce{AgNO3})=107.87\space g/mol + 14.01\space g/mol+3\times16.00\space g/mol = 169.88\space g/mol$. Given the mass of $\ce{AgNO3}$, $m = 5.0\space g$. Using the formula $n=\frac{m}{M}$, the number of moles of $\ce{AgNO3}$ is: $$n(\ce{AgNO3})=\frac{5.0\space g}{169.88\space g/mol}\approx0.0294\space mol$$

Step3: Determine moles of $\ce{BaCl2}$ from stoichiometry

From the balanced equation, the mole ratio of $\ce{AgNO3}$ to $\ce{BaCl2}$ is $2:1$. So, $n(\ce{BaCl2})=\frac{1}{2}n(\ce{AgNO3})$. Substituting the value of $n(\ce{AgNO3})$: $$n(\ce{BaCl2})=\frac{1}{2}\times0.0294\space mol = 0.0147\space mol$$

Step4: Calculate mass of $\ce{BaCl2}$

The molar mass of $\ce{BaCl2}$ is $M(\ce{BaCl2}) = 137.33\space g/mol+2\times35.45\space g/mol=208.23\space g/mol$. Using the formula $m = n\times M$, the mass of $\ce{BaCl2}$ is: $$m(\ce{BaCl2})=0.0147\space mol\times208.23\space g/mol\approx3.06\space g$$

Answer:

$\approx3.1$ (or more precisely $\approx3.06$)