According to the following reaction, how many grams of water are produced in the complete reaction of 31.8 grams of barium hydroxide?
ba(oh)2(aq) + h2so4(aq) --> baso4(s) + 2h2o(l)

6.696 g of water (H₂O)

Explanation:

First we calculate the number of moles of barium hydroxide:

number of moles = mass / molecular weight

number of moles = 31.8 / 171.3 = 0.186 moles of barium hydroxide

Form the chemical reaction we may devise the following reasoning:

if         1 mole of barium hydroxide produces 2 moles of water

then   0.186 moles of barium hydroxide produces X moles of water

X = (0.186 × 2) / 1 = 0.372 moles of water

mass = number of moles × molecular weight

mass of water = 0.372 × 18 = 6.696 g

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