The mass of aluminum hydroxide formed during the interaction of the solution contains 25 g of sodium

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.



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