1. Let’s write down the reaction equation:
2Al (OH) 3 + 3H2CO3 = Al2 (CO3) 3 + 6H2O.
2. Find the amount of acid and base:
n (H2CO3) = m (H2CO3) / M (H2CO3) = 50 g / 62 g / mol = 0.81 mol.
n (Al (OH) 3) = m (Al (OH) 3) / M (Al (OH) 3) = 100 g / 78 g / mol = 1.28 mol.
3. According to the reaction equation, it can be seen that aluminum hydroxide is in short supply, therefore, from it we find the amount of water, and then the mass:
2n (Al (OH) 3) = 6n (H2O)
n (H2O) = 1 / 3n (Al (OH) 3) = 1/3 * 0.81 mol = 0.27 mol.
m (H2O) = n (H2O) * M (H2O) = 0.27 mol * 18 g / mol = 4.84 g.
Answer: m (H2O) = 4.84 g.
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