## Some problems I just don't get.

**Anonymous**

What mass of C6H13BR will be produced by a reaction giving 65% yield if 12.5mL of liquid C6H12 (density = 0.673g/mL) is treated with 2.70L HBr (g) at STP?

C6H12 + HBr ----> C6H13Br

I don't understand what it means by density. Should I just work the problem by finiding the limiting reactant and caculate the theoretical yield and multiply it by a percentage?

Anonymous| 26 January 2004 - 6:42amWell you need to calculate the limiting reagent. The reason they gave you density and a volume was so you can find the mass of C6H12 and from that figure out the moles.

Anonymous| 30 January 2004 - 5:28pm[quote]What mass of C6H13BR will be produced by a reaction giving 65% yield if 12.5mL of liquid C6H12 (density = 0.673g/mL) is treated with 2.70L HBr (g) at STP?

C6H12 + HBr ----> C6H13Br

I don't understand what it means by density. Should I just work the problem by finiding the limiting reactant and caculate the theoretical yield and multiply it by a percentage?[/quote]

Density is defined by mass/volume.

You can find the limiting agent by converting to moles. Remember, whenever you see a chemical equation you will always need convert to moles (with few exceptions).

To find the moles of C6H12, multiply the volume by density. Notice that the volume cancells out to give you the grams. Find the molar mass of the above compound and thus find the number moles (since you have the actual grams).

Notice that the HBr is a gas at standard temperature and pressure. One mole of any gas at standard temperature occupying a volume of 22.4L will exert standard pressure (1 atm). Thus 1moles/22.4L, you can use this as a conversion factor to find the moles e.g. 2.70L(1mole/22.4L)=?

I am sure you are familiar with finding the limiting reagent. If you need help, just tell me. Find the limiting reagent, find the number of moles of the product yielded. This is the theoretical yield. Multiply by .65 (65%).

Hope this helps.