# Question 1bb2b

Mar 15, 2015

You would see $\text{12.7}$ grams of silver chloride and $\text{7.51 g}$ of sodium nitrate formed after this reaction takes place.

So, you're basically dealing with a double replacement reaction which leads to the formation of a white precipitate, silver chloride. The balanced chemical equation for this reaction looks like this

$A g N {O}_{3 \left(a q\right)} + N a C {l}_{\left(a q\right)} \to A g C {l}_{\left(s\right)} + N a N {O}_{3 \left(a q\right)}$

Notice the $\text{1:1}$ mole ratio that exists between all the compounds involved in this reaction - 1 mole of silver nitrate needs 1 mole of sodium chloride to produce 1 mole of silver chloride and 1 mole of sodium nitrate.

In an ideal scenario, you'll have enough moles of both reactants so that neither of them can act as a limiting reagent. You can determine if this is the case by calculating how many moles of each reactant you get - use their molar masses

${\text{15.0 g AgNO"_3 * "1 mole AgNO"_3/"169.87 g" = "0.0883 moles AgNO}}_{3}$

$\text{15.0 moles NaCl" * "1 mole NaCl"/"58.443 g" = "0.257 moles NaCl}$

This shows that you have insufficient moles of silver nitrate to allow for all the moles of sodium chloride to react $\to$silver nitrate is a limiting reagent..

Since only 0.0883 moles of each reactant will actually react, the number of moles of silver chloride and sodium nitrate produced will be equal to 0.0883.

The masses produced will be

$\text{0.0883 moles AgCl" * "143.32 g"/"1 mole" = "12.7 g AgCl}$

${\text{0.0883 moles NaNO"_3 * "85.0 g"/"1 mole" = "7.51 g NaNO}}_{3}$

The total mass produced will be

m_("total") = m_(AgCl) + m_(NaNO_3) = 12.7 + 7.51 = "20.2 g"

This is equal to the total mass that reacted, which is

${m}_{\text{reacted") = m_(AgNO_3) + m_("NaCl - reacted}}$
m_("reacted") = 15.0 + 5.16 = "20.2 g"#

Keep in mind that this is your theoretical yield - the amount produced when the reaction has a 100% yield.