Respuesta :
The balanced chemical equation for reaction of [tex]AgNO_{3}[/tex] and [tex]Na_{2}SO_{4}[/tex] is as follows:
[tex]2 AgNO_{3}+Na_{2}SO_{4}\rightarrow 2NaNO_{3}+Ag_{2}SO_{4}[/tex]
From the balanced chemical equation, 2 mol of [tex]AgNO_{3}[/tex] reacts with 1 mol of [tex]NaNO_{3}[/tex].
First calculating number of moles of [tex]NaNO_{3}[/tex] as follows:
[tex]M=\frac{n}{V}[/tex]
On rearranging,
[tex]n=M\times V[/tex]
Here, M is molarity and V is volume. The molarity of [tex]NaNO_{3}[/tex] is given 0.274 M or mol/L and volume 155 mL, putting the values,
[tex]n=0.274 mol/L\times 155\times 10^{-3}mL=0.04247 mol[/tex]
Since, 1 mol of [tex]NaNO_{3}[/tex] reacts with 2 mol of [tex]AgNO_{3}[/tex] thus, number of moles of [tex]AgNO_{3}[/tex] will be [tex]2\times 0.04247 mol=0.08494 mol[/tex].
Now, molarity of [tex]AgNO_{3}[/tex] is given 0.305 M or mol/L thus, volume can be calculated as follows:
[tex]V=\frac{n}{M}=\frac{0.08494 mol}{0.305 mol/L}=0.2785 L=278.5 mL[/tex]
Therefore, volume of [tex]AgNO_{3}[/tex] is 278.5 mL.
155.0 mL of 0.274 M Na₂SO₄ react exactly with 279 mL of 0.305 M AgNO₃.
What is a balanced equation?
A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products.
Let's consider the balanced equation for the reaction between AgNO₃ and Na₂SO₄.
2 AgNO₃ + Na₂SO₄ ⇒ 2 NaNO₃ + Ag₂SO₄
- Step 1. Calculate the reacting moles of Na₂SO₄.
155.0 mL of 0.274 M Na₂SO₄ react.
0.1550 L × 0.274 mol/L = 0.0425 mol
- Step 2. Calculate the reacting moles of AgNO₃.
The molar ratio of AgNO₃ to Na₂SO₄ is 2:1.
0.0425 mol Na₂SO₄ × (2 mol AgNO₃/1 mol Na₂SO₄) = 0.0850 mol AgNO₃
- Step 3. Calculate the volume that contains 0.0850 moles of AgNO₃
The concentration of AgNO₃ is 0.305 M.
0.0850 mol × (1 L/0.305 mol) = 0.279 L = 279 mL
155.0 mL of 0.274 M Na₂SO₄ react exactly with 279 mL of 0.305 M AgNO₃.
Learn more about stoichiometry here: https://brainly.com/question/16060223
