How many moles of iron(lll) sulfide, Fe2S3, would be produced from the complete reaction of 449 g iron(lll) bromide, FeBr3?

Chemistry

1 answer:

SpyIntel [72]3 years ago

7 0

Answer:

158.004 Grams of Fe2S3

Explanation:

iron(lll) sulfide = Fe2S3 = Molar mass: 207.9 g/mol

iron(lll) bromide = FeBr3 = Molar mass: 295.56 g/mol

Since we don't have the chemical equation we will have to make one I guess.

2 FeBr3 + 3 Na2S -----> 6 NaBr + Fe2S3

449 g iron(lll) bromide = FeBr3

So we have 449/295.56 = moles of FeBr3 = 1.52 Moles

So ratio is 2:1 FeBr3 to Fe2S3

So we need 0.76 moles of Fe2S3

0.76 x 207.9 = 158.004Grams

You might be interested in

A student carried out a simple distillation on a compound known to boil at 124 oc and they reported a boiling point of 116-117 o

chubhunter [2.5K]

Answer :

The correct answer is due to incorrect position of thermometer .

Distillation is process of separating two volatile liquids on the basis of their boiling points .In involves the following set up ( shown in image )

Since the Boiling point is found is 116-117 C which is near to 124 C , also the purity was checked and thermometer was calibrated .

So the only error could be possible is position of thermometer .

The position of bulb of thermometer is very important to get an accurate reading of boiling point .The vapor so formed should touch the side walls of bulb as soon as they are formed .

May be the bulb of thermometer was kept far from the vapors formed by liquid so, some of the vapor before reaching to bulb start condensing and their Temperature decreases .

Hence , the temperature shown by thermometer was decrease than actual boiling point due to incorrect position of thermometer.