If you start with ten grams of lithium hydroxide, how many grams of lithium bromide will be produced? Round to the nearest tenth. Don't forget the units.
You don't have a reaction, only a question. What is the reaction?
To determine how many grams of lithium bromide will be produced from ten grams of lithium hydroxide, we need to use the balanced chemical equation of the reaction. The balanced equation for the reaction between lithium hydroxide and lithium bromide is:
2 LiOH + LiBr → Li2O + 2 H2O + LiBr
From the balanced equation, we can see that for every 2 moles of lithium hydroxide (LiOH), we get 1 mole of lithium bromide (LiBr) produced.
To calculate the number of moles of lithium hydroxide, we use the formula:
moles = mass / molar mass
The molar mass of lithium hydroxide (LiOH) is:
(1 mol Li x 6.94 g/mol) + (1 mol O x 16.00 g/mol) + (1 mol H x 1.01 g/mol) = 23.95 g/mol
So, for ten grams of lithium hydroxide:
moles of LiOH = 10 g / 23.95 g/mol ≈ 0.418 mol
Now, we can use the stoichiometry of the balanced equation to find the moles of lithium bromide produced. From the balanced equation, we can see that for every 2 moles of lithium hydroxide, we get 1 mole of lithium bromide produced.
The moles of lithium bromide produced can be calculated as:
moles of LiBr = (moles of LiOH / 2) ≈ (0.418 mol / 2) ≈ 0.209 mol
Finally, to convert moles to grams, we multiply the moles by the molar mass of lithium bromide (LiBr):
molar mass of LiBr = (1 mol Li x 6.94 g/mol) + (1 mol Br x 79.90 g/mol) ≈ 86.84 g/mol
grams of LiBr = moles of LiBr × molar mass of LiBr
≈ 0.209 mol × 86.84 g/mol ≈ 18.15 g
Therefore, approximately 18.15 grams of lithium bromide will be produced from ten grams of lithium hydroxide.