Stoichiometric reactions with weight percent

AI Thread Summary
The discussion revolves around calculating the grams of 6.000 wt% aqueous HF needed to provide a 50% excess for a reaction with 10.0 mL of 0.0236 M Th4+. The initial calculations indicate that the molarity of Th(4+) is 2.36*10^-4, leading to a requirement of 9.44*10^-4 moles of F-. After accounting for the 50% excess, the total grams of HF needed is calculated to be 0.376g, resulting in 0.627g when considering the weight percent. A clarification is made that 50% excess differs from 100% excess, which is crucial for accurate calculations. Understanding the distinction between these excess percentages is essential for correct stoichiometric calculations.
emc92
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i can't figure out how to calculate this. Please help!

How many grams of 6.000 wt% aqueous HF are required to provide a 50% excess to react with 10.0 mL of 0.0236 M Th4+ by the following reaction?

Th(4+) + 4F - → ThF4 (s)

I found the molarity for Th(4+) to be 2.36*10^-4.. I'm not really sure where to go from here =/
 
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How many moles of F- do you need? What mass of HF will contain this number of moles?
 
i tried to solve it again..
i came up with 0.627g. is this right?
 
Show how you got it.
 
using the moles of Th, i multiplied that by 4 to find the number of moles of HF, which turns out o be 9.44*10^-4. Then i changed it to grams, giving me 1.88*10^-2g. since it said 50% excess, the total grams of HF would be 0.376g. then 6 %wt would be 0.627g
 
emc92 said:
using the moles of Th, i multiplied that by 4 to find the number of moles of HF, which turns out o be 9.44*10^-4. Then i changed it to grams, giving me 1.88*10^-2g.

OK up to now.

since it said 50% excess, the total grams of HF would be 0.376g. then 6 %wt would be 0.627g

You ate zero, but I think I know what you did. 50% excess and 100% excess are two different things.
 
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