Tin(IV) sulfide, SnS 2 , a yellow pigment, can be produced using the following reaction. SnBr 4 ( aq ) + 2 Na 2 S ( aq ) ⟶ 4 NaBr ( aq ) + SnS 2 ( s ) Suppose a student adds 47.7 mL of a 0.474 M solution of SnBr 4 to 43.4 mL of a 0.179 M solution of Na 2 S . Identify the limiting reactant.

The limiting reactant is Na2S.

We have the equation of the reaction as follows;

SnBr4 ( aq ) + 2Na2S ( aq ) ⟶ 4NaBr ( aq ) + SnS2 ( s )

Number of moles of SnBr4 = concentration × volume

= 0.474 M ×  47.7/1000 L = 0.023 moles

Number of moles of Na2S = 0.179 M  × 43.4/1000 L = 0.0078 moles

From the reaction equation;

1 mole of SnBr4 reacts with 2 moles of Na2S

0.023 moles of SnBr4 reacts with 0.023 moles × 2 moles/1 mole

= 0.046 moles

We can see that the amount of Na2S required for the reaction is far greater than what is available hence there is not enough Na2S present. Na2S is the limiting reactant.

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is the limiting reagent

Explanation:

To calculate the number of moles for given molarity, we use the equation:

    .....(1)

1. Molarity of solution = 0.474 M

Volume of solution = 47.7 mL = 0.0477 L

Putting values in equation 1, we get:

2. Molarity of solution = 0.179 M

Volume of solution = 43.4 mL = 0.0434 L

Putting values in equation 1, we get:

According to stoichiometry :

2 moles of require 1 mole of

Thus 0.00777 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.