Chemical Forums

Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Nori on September 01, 2019, 12:11:25 PM

Title: Percentage Yield
Post by: Nori on September 01, 2019, 12:11:25 PM
I'm having a lot of trouble with this for some reason, can someone give me some hint or step to solve this?

When 2.36 moles of iron iii bromide yields 6.14 moles of sodium bromide, what is the percentage yield of the sodium bromide?

2FeB3+3Na2S --> Fe2S3+6NaBr

What I did was convert the moles into g and I don't know what to do next
Title: Re: Percentage Yield
Post by: AWK on September 01, 2019, 12:56:00 PM
FeB3 - printing error

Quote
What I did was convert the moles into g

You can make calculations with moles.
How many moles of NaBr should be formed from 2.36 moles of iron(III) bromide - this is a theoretical yield.
Title: Re: Percentage Yield
Post by: Vidya on September 02, 2019, 01:22:42 AM
Quote from: Nori on September 01, 2019, 12:11:25 PM


What I did was convert the moles into g and I don't know what to do next
Calculate moles formed as suggested by AWK and take ratio of moles formed with theoretical moles multiplied by 100%. No need of converting in grams.