Question

`BaCO_3 + C(s) + H_2O(l) -> CO(g) + Ba(OH)_2`. If 394 g of `BaCO_3` are reacted with excess carbon and water, what mass of `Ba(OH)_2` should be produced?

Answer 1

You should expect 342.6 g of `Ba(OH)_2`

Explanation:

`394 g of BaCO_3` is 2.00 moles.

`(394 g)/(197.3 (g/"mol")) = 2.00` mol

Looking at the equation, you see that the coefficient of both `BaCO_3` and `Ba(OH)_2` is a one. So, the amount of `Ba(OH)_2` produced will be equal to the amount of `BaCO_3` reacted.

`2.00 " mol " Ba(OH)_2 xx 171.3 (g/"mol") = 342.6 g`