# Calculate the mass of the precipitate precipitated by the interaction of 700 g of a 10% solution of Al (NO3) 3

Calculate the mass of the precipitate precipitated by the interaction of 700 g of a 10% solution of Al (NO3) 3 with the required amount of BaOH. What is the amount of the precipitate formed?

1.write the exchange reaction for aluminum nitrate and barium hydroxide, since all nitrates are soluble, we get a precipitate of aluminum hydroxide:

2Al (NO3) 3 + 3Ba (OH) 2 = 2Al (OH) 3 ↓ + 3Ba (NO3) 2;

2.m (Al (NO3) 3) = m (solution) * w ((Al (NO3) 3));

m (Al (NO3) 3) = 700 * 0.1 = 70 g;

n (Al (NO3) 3) = m (Al (NO3) 3): M (Al (NO3) 3);

M (Al (NO3) 3) = 27 + 14 * 3 + 16 * 3 * 3 = 213 g / mol;

n (Al (NO3) 3) = 70: 213 = 0.3286 mol;

3. find the chemical amount of the resulting sediment:

n (Al (OH) 3) = n (Al (NO3) 3) = 0.3286 mol;

4.Calculate the mass of aluminum hydroxide:

m (Al (OH) 3) = n (Al (OH) 3) * M (Al (OH) 3);

M (Al (OH) 3) = 27 + 16 * 3 + 1 * 3 = 78 g / mol;

m (Al (OH) 3) = 0.3286 * 78 = 25.63 g.