If 25g of lithium reacts with selenium to from 75g of lithium selenide, how many grams of selenium will react?
To find out how many grams of selenium will react with 25g of lithium, we first need to calculate the molar mass of lithium selenide.
The molar mass of lithium is approximately 6.94 g/mol, and the molar mass of selenium is approximately 78.97 g/mol. Since lithium selenide has a 1:1 ratio of lithium to selenium, the molar mass of lithium selenide is the sum of the molar masses of lithium and selenium:
6.94 g/mol (lithium) + 78.97 g/mol (selenium) = 85.91 g/mol (lithium selenide)
Next, we can calculate the number of moles of lithium selenide formed from 25g of lithium:
25g / 6.94 g/mol = 3.60 mol of lithium
Since the molar ratio of lithium to selenium in lithium selenide is 1:1, there are also 3.60 mol of selenium in 3.60 mol of lithium selenide.
Finally, we can convert the moles of selenium to grams:
3.60 mol * 78.97 g/mol = 284.29 g
Therefore, 284.29 grams of selenium will react with 25g of lithium to form 75g of lithium selenide.