Respuesta :
[tex]SnBr_{4(aq)}+2Na_2S_{(aq)}\rightarrow4NaBr_{(aq)}+SnS_{2(s)}[/tex]
Given:
Concentration of SnBr4=0.478 M
Volume of SnBr4=48.1 mL
Concentration of Na2S= 0.160 M
Volume of Na2S= 48.8 mL
We will convert the molarity of the reactants into moles:
For SnBr4: if 0.478 mole is in 1000 mL then x mole is in 48.1 mL:
[tex]\begin{gathered} x=\frac{0.478mol\times48.1mL}{1000mL} \\ x=0.023mol \end{gathered}[/tex]For Na2S: if 0.160 mol is in 1000mL then x mole is in 48.8mL:
[tex]\begin{gathered} x=\frac{0.160mol\times48.8mL}{1000mL} \\ x=0.008mol \\ \end{gathered}[/tex]No we are able to calculate the limiting reactant by using conversion factors.
A)To use all of the SnBr4, how many moles of Na2S do we need:
[tex]\begin{gathered} 0.023mol\text{ }SnBr_4\times\frac{2mol\text{ }Na_2S}{1mol\text{ }SnBr_4}=0.046mole\text{ }Na_2S \\ \end{gathered}[/tex]B) To use all of the Na2S, how many moles of SnBr4 do we need:
[tex]0.008mol\text{ }Na_2S\times\frac{1mol\text{ }SnBr_4}{2mol\text{ }Na_2S}=0.004mol\text{ }SnBr_4[/tex]Answer: Limiting reactant is Na2S.
Explanation: We need 0.046 mole of Na2S to use all of the SnBr4 and we only have 0.008 mole. We need 0.004 mole of SnBr4 to use all of the Na2S and we have 0.023 mole. This means Na2S is the limiting reactant. The limiting reactant determines the amount of product that can be formed.
Now we will calculate the theoretical yield, To determine this we will use the limiting reactant to do so because this determine the amount of product that can be formed.
[tex]0.008\text{ }molNa_2S\times\frac{1mol\text{ }SnS_2}{2mol\text{ }Na_2S}=0.004mol\text{ }SnS_2[/tex]Answer: The theoretical yield of tin(iv) sulfide in moles is 0.004.
