# 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?

$M M C {a}_{3} {\left(B {O}_{3}\right)}_{2} = 40 \times 3 + \left(10 , 8 + 16 \times 3\right) \times 2 = 120 + 117 , 6 = 237 , 6 \frac{g}{m o l}$
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 $10 g \times \frac{120}{237.6} = 5.05 g$