Answer:
e volume of the 1.22 M aqueous sodium sulfide solution needed to react with 2.75 g of bismuth (III) bromide is 5.75 mL.
Step-by-step explanation:
The balanced chemical equation for the reaction between sodium sulfide and bismuth (III) bromide is:
Na2S(aq) + BiBr3(aq) → 3NaBr(aq) + Bi2S3(s)
From the balanced equation, we can see that one mole of sodium sulfide reacts with one mole of bismuth (III) bromide to produce two moles of sodium bromide and one mole of bismuth sulfide.
First, we need to calculate the number of moles of bismuth (III) bromide in 2.75 g:
molar mass of BiBr3 = 391.69 g/mol
moles of BiBr3 = mass/molar mass = 2.75 g/391.69 g/mol = 0.00701 mol
Since the reaction is 1:1 between sodium sulfide and bismuth (III) bromide, we need 0.00701 moles of Na2S.
To calculate the volume of the sodium sulfide solution needed, we can use the equation:
Molarity = moles/volume (in liters)
Rearranging, we get:
Volume (in liters) = moles/Molarity
We have the moles and the Molarity is given as 1.22 M, so we can plug in the values:
Volume (in liters) = 0.00701 mol/1.22 M = 0.00575 L
Finally, we convert the volume from liters to milliliters:
Volume (in mL) = 0.00575 L x 1000 mL/L = 5.75 mL
Therefore, the volume of the 1.22 M aqueous sodium sulfide solution needed to react with 2.75 g of bismuth (III) bromide is 5.75 mL.