Determine the mass of the precipitate that forms when 166 g of sodium carbonate interacts with magnesium chloride.

Let’s execute the solution:
1. Let’s compose an equation according to the condition of the problem:
m = 166 g. X g. -?
Na2CO3 + MgCl2 = MgCO3 + 2NaCl – ion exchange, magnesium carbonate was obtained in the form of a white precipitate;
2. Let’s make the calculations:
M (Na2CO3) = 105.8 g / mol;
M (MgCO3) = 84.3 g / mol.
3. Determine the amount of the original substance:
Y (Na2CO3) = m / M = 166 / 105.8 = 1.57 mol;
Y (MgCO3) = 1.57 mol since the amount of these substances according to the equation is 1 mol.
4. Find the mass of the product:
m (MgCO3) = Y * 1.57 * 84.3 = 132.35 g.
Answer: magnesium carbonate with a mass of 132.35 g was formed.

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