Question

6 Li + Ca3(BO3)2 ---> 3 Ca + 2 Li3BO3 How many grams of calcium metal would be produced if you started the reaction with 10 grams of calcium borate?

How much of the excess reactant would be left over at the end of the reaction?

Answer 1

5,05 g

Explanation:

`MM Ca_3(BO_3)_2= 40 xx 3 + (10,8 + 16 xx3) xx 2 = 120 + 117,6 =237,6 g/(mol)`
You have to do a proportion: from 1 mol of calcium borate (237,6 g) you obtain 3 mol of calcium atoms (120 g), so from 10 g of calcium borate you obtain `10 g xx 120/237.6=5.05 g`