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How many grams of [tex]AgCl[/tex] (molar mass = [tex]143.5 \, g/mol[/tex]) can be produced when [tex]25.0 \, ml[/tex] of [tex]1.5M \, AgNO_3[/tex] completely reacts with excess [tex]NaCl(s)[/tex]?

[tex]AgNO_3(aq) + NaCl(s) \rightarrow AgCl(s) + NaNO_3(aq)[/tex]

Answer :

Final answer:

The reaction of 25.0ml of 1.5M AgNO3 with excess NaCl can produce 5.38 grams of AgCl.

Explanation:

The question is about a chemical reaction in which silver nitrate (AgNO3) reacts with sodium chloride (NaCl) to produce silver chloride (AgCl) and sodium nitrate (NaNO3). The molar concentration of AgNO3 is given as 1.5M and the volume is 25.0ml.

First, we need to convert the volume of AgNO3 from milliliters to liters because molar concentration is defined in terms of moles per liter. So, 25.0ml is equivalent to 0.025L. Then, multiply the volume of the AgNO3 solution by its molar concentration to find the number of moles of AgNO3: (0.025 L)*(1.5 mol/L) = 0.0375 mol AgNO3.

The chemical equation shows that one mole of AgNO3 produces one mole of AgCl. Therefore, 0.0375 mol of AgNO3 will produce 0.0375 mol of AgCl. Using the molar mass of AgCl (143.5g/mol), we can convert moles of AgCl to grams: (0.0375 mol)*(143.5 g/mol) = 5.38 g AgCl.

In conclusion, 5.38 grams of AgCl can be produced when 25.0ml of 1.5M AgNO3 completely reacts with excess NaCl.

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