What mass of magnesium nitrate is formed by the action of 8 grams of magnesium oxide with nitric acid, if the magnesium oxide contains 15% impurities?
Let’s execute the solution:
1. According to the condition of the problem, we write the equation:
m = 8 g; 15% impurity. X g -?
MgO + 2HNO3 = Mg (NO3) 2 + H2O – exchange, magnesium nitrate was obtained;
M (MgO) = 40.3 g / mol.
M Mg (NO3) 2 = 148.3 g / mol.
3. Determine the mass of the original substance without impurity, its amount:
m (MgO) = 8 (1-0.15) = 6.8 g;
Y (MgO) = m / M = 6.8 / 40.3 = 0.17 mol;
Y Mg (NO3) 2 = 0.17 mol since the amount of substances according to the equation is 1 mol.
4. Find the mass of the product:
m Mg (NO3) 2 = Y * M = 0.17 * 148.3 = 25.2 g.
Answer: obtained magnesium nitrate weighing 25.2 g.
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