The mass of aluminum hydroxide formed during the interaction of the solution contains 25 g of sodium hydroxide and 26.7 g of aluminum chloride is equal.
1. Let us write down the equation of the ongoing reaction between sodium hydroxide and aluminum chloride:
3NaOH + AlCl3 = Al (OH) 3 + 3NaCl;
2.Calculate the chemical quantities of reacting substances:
n (NaOH) = m (NaOH): M (NaOH);
M (NaOH) = 23 + 17 = 40 g / mol;
n (NaOH) = 25:40 = 0.625 mol;
n (AlCl3) = m (AlCl3): M (AlCl3);
M (AlCl3) = 27 + 35.5 * 3 = 133.5 g / mol;
n (AlCl3) = 26.7: 133.5 = 0.2 mol;
3.Aluminum chloride was taken in a slight deficiency, we determine the amount of the formed hydroxide:
n (Al (OH) 3) = n (AlCl3) = 0.2 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 + 3 * 17 = 78 g / mol;
m (Al (OH) 3) = 0.2 * 78 = 15.6 g.
Answer: 15.6 g.