Answer: [tex]Na_2S[/tex] is the limiting reagent
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}[/tex] .....(1)
1. Molarity of [tex]SnBr_4[/tex] solution = 0.474 M
Volume of [tex]SnBr_4[/tex] solution = 47.7 mL = 0.0477 L
Putting values in equation 1, we get:
[tex]\text{Moles of} SnBr_4={0.474}\times{0.0477}=0.0226moles[/tex]
2. Molarity of [tex]Na_2S[/tex] solution = 0.179 M
Volume of [tex]Na_2S[/tex] solution = 43.4 mL = 0.0434 L
Putting values in equation 1, we get:
[tex]\text{Moles of} Na_2S={0.179}\times{0.0434}=0.00777moles[/tex]
[tex]SnBr_4(aq)+2Na_2S(aq)\rightarrow 4NaBr(aq)+SnS_2(s)[/tex]
According to stoichiometry :
2 moles of [tex]Na_2S[/tex] require 1 mole of [tex]SnBr_4[/tex]
Thus 0.00777 moles of [tex]Na_2S[/tex] will require=[tex]\frac{1}{2}\times 0.00777=0.00388moles[/tex] of [tex]SnBr_4[/tex]
Thus [tex]Na_2S[/tex] is the limiting reagent as it limits the formation of product and [tex]SnBr_4[/tex] is the excess reagent.
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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