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omgneeky · May 4, 2009, 12:48 PM

Zn+2HBR--->ZnBr2+H2

86.4 grams of Zn reacts with how many grams of HBr?

Perito · May 4, 2009, 02:22 PM

Originally Posted by omgneeky

$Zn+2HBr \, \rightarrow \, ZnBr_2+H_2$

86.4 grams of Zn reacts with how many grams of HBr?

$grams\,of\,Zn \,\div\,\frac {grams}{mole}\,of\,Zn\,=\,moles\,of\,Zn$

grams/mole of zinc is the atomic weight of zinc. You can look that up.

The balanced equation shows that you need two moles of HBr to react with one mole of zinc, so multiply the number of moles by 2

$moles\,of\,HBr\,=\,moles\,of\,Zn\,\times\,2$

$moles\,of\,HBr\,\times\,\frac {grams\,of\,HBr}{moles\,of\,HBr}\,=\,grams\,of\,Br$

Note how you can multiply units to figure out what to do.

The molecular weight of HBr is "grams of HBr/Mole of HBr". You can look that up or calculate that also. Once you multiply that out, you'll have your answer.