LiOH+HBr---> LiBr +h20 If you start with 10.0 grams of lithium hydroxide, how many grams of lithium bromide will be produced?

LiOH+HBr---> LiBr +h20. Moles of LiOH = 10/24 = 0.41moles. According to stoichiometry, moles of LiOH = moles of LiBr = 0.41moles. Therefore mass of LiBr =moles of LiBr x molecular weight of LiBr = o.41 x 87 = 35.67g. Hope it helps

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CintiaSalazar CintiaSalazar · Answer 2 · 2021-09-24T15:30:49+02:00

Taking into account the stoichiometry of the reaction, 36.25 grams of lithium bromide will be produced.

First of all, the balanced reaction is:

LiOH + HBr → LiBr + H₂O

By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:

LiOH: 1 mole
HBr: 1 mole
LiBr: 1 mole
H₂O: 1 mole

The molar mass of each compound is:

LiOH: 24 g/mole
HBr: 81 g/mole
LiBr: 87 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

LiOH: 1 mole×24 g/mole= 24 g
HBr: 1 mole×81 g/mole= 81 g
LiBr: 1 mole×87 g/mole= 87 g
H₂O: 1 mole×18 g/mole= 18 g

So, you can apply the following rule of three: if by stoichiometry 24 g of LiOH react to form 87 g of LiBr, 10 g of LiOH form how much mass of LiBr?

[tex]mass of LiBr=\frac{10 grams of LiOHx87gramsof LiBr}{24 grams of LiOH}[/tex]

mass of LiBr= 36.25 grams of LiBr

Finally, 36.25 grams of lithium bromide will be produced.

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