AgNO3 (3 is a subscript) + NaCl —> NaNO3 (3is a subscript) + AgCl. If we use 15.0 grams of the first reactant and 15.0 grams of the second reactant, how many total grams of the definite product should we theoretically see at the end of the experiment?

You would see ##”12.7″## grams of silver chloride and ##”7.51 g”## of sodium nitrate formed after this reaction takes place.

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

##AgNO_(3(aq)) + NaCl_((aq)) -> AgCl_((s)) + NaNO_(3(aq))##

Notice the ##”1:1″## that exists between all the 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 . You can determine if this is the case by calculating how many moles of each reactant you get – use their molar masses

##”15.0 g AgNO”_3 * “1 mole AgNO”_3/”169.87 g” = “0.0883 moles AgNO”_3##

##”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 ##->##silver nitrate is a ..

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

##”0.0883 moles AgCl” * “143.32 g””1 mole” = “12.7 g AgCl”##

##”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_(“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% .

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