Calculate the mass of the precipitate obtained by reacting 160 g of sodium hydroxide (NaOH)
April 18, 2021 | education
| Calculate the mass of the precipitate obtained by reacting 160 g of sodium hydroxide (NaOH) with magnesium chloride (MgCl2).
Let’s write the reaction equation:
MgCl2 + 2NaOH = Mg (OH) 2 ↓ + 2NaCl
Find the amount of sodium hydroxide substance:
v (NaOH) = m (NaOH) / M (NaOH) = 160/40 = 4 (mol).
According to the reaction equation, 1 mol of Mg (OH) 2 is formed per 2 mol of NaOH, therefore:
v (Mg (OH) 2) = v (NaOH) / 2 = 4/2 = 2 (mol).
Thus, the mass of the obtained magnesium hydroxide precipitate is:
m (Mg (OH) 2) = v (Mg (OH) 2) * M (Mg (OH) 2) = 2 * 58 = 116 (g).
Answer: 116 g.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.