What is the mass of the precipitate obtained by the reaction of 345 g of a 10% solution

| June 23, 2021 | education

What is the mass of the precipitate obtained by the reaction of 345 g of a 10% solution of zinc nitrate with an excess of sodium hydroxide?

1. The interaction of zinc nitrate with sodium hydroxide proceeds according to the equation:

Zn (NO3) 2 + 2NaOH = Zn (OH) 2 ↓ + 2NaNO3;

2. Let’s calculate the mass of zinc nitrate:

m (Zn (NO3) 2) = w (Zn (NO3) 2) * m (solution) = 0.1 * 345 = 34.5 g;

3. Calculate the chemical amount of zinc nitrate:

n (Zn (NO3) 2) = m (Zn (NO3) 2): M (Zn (NO3) 2);

M (Zn (NO3) 2) = 65 + 2 * 14 + 6 * 16 = 189 g / mol;

n (Zn (NO3) 2) = 34.5: 189 = 0.1825 mol;

4. Determine the amount of zinc hydroxide obtained:

n (Zn (OH) 2) = n (Zn (NO3) 2) = 0.1825 mol;

5. Find the mass of the sediment:

m (Zn (OH) 2) = n (Zn (OH) 2) * M (Zn (OH) 2);

M (Zn (OH) 2) = 65 + 2 * 17 = 99 g / mol;

m (Zn (OH) 2) = 0.1825 * 99 = 18.07 g.

Answer: 18.07 g.



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