Final answer:
To find the mass of magnesium nitrate produced in the reaction between magnesium carbonate and nitric acid, we can use stoichiometry and the concept of limiting reactants. We can determine the mass of magnesium nitrate, magnesium carbonate, and nitric acid remaining after the reaction is complete.
Step-by-step explanation:
To find the mass of magnesium nitrate produced, we need to use stoichiometry and the concept of limiting reactants. The balanced chemical equation is:
MgCO3(s) + 2HNO3(aq) ⟶ Mg(NO3)2(aq) + H2O(l) + CO2(g)
From the equation, we can see that 1 mole of magnesium carbonate (MgCO3) reacts with 2 moles of nitric acid (HNO3) to produce 1 mole of magnesium nitrate (Mg(NO3)2).
We can use this information to calculate the moles of magnesium nitrate produced:
31.0 g of magnesium carbonate is equal to 31.0 g / 84.31 g/mol = 0.367 mol of magnesium carbonate.
According to the stoichiometry of the balanced equation, the moles of magnesium nitrate produced will be half of the moles of magnesium carbonate.
Therefore, the moles of magnesium nitrate produced will be 0.367 mol * (1/2) = 0.1835 mol.
The molar mass of magnesium nitrate (Mg(NO3)2) is 148.31 g/mol.
Therefore, the mass of magnesium nitrate produced will be 0.1835 mol * 148.31 g/mol = 27.20 g.
To calculate the mass of magnesium carbonate remaining after the reaction is complete, we can use the stoichiometry of the balanced equation.
Since 1 mole of magnesium carbonate reacts with 2 moles of nitric acid to produce 1 mole of magnesium nitrate, the moles of magnesium carbonate that reacted will be half of the moles of magnesium nitrate produced, which is 0.1835 mol.
The molar mass of magnesium carbonate (MgCO3) is 84.31 g/mol.
Therefore, the mass of magnesium carbonate remaining will be 0.1835 mol * 84.31 g/mol = 15.46 g.
To calculate the mass of nitric acid remaining after the reaction is complete, we need to subtract the moles of nitric acid that reacted from the initial moles of nitric acid. The initial moles of nitric acid can be calculated using its molar mass:
15.0 g of nitric acid is equal to 15.0 g / 63.01 g/mol = 0.2381 mol of nitric acid.
Since 1 mole of magnesium carbonate reacts with 2 moles of nitric acid, the moles of nitric acid that reacted will be 0.367 mol * 2 = 0.734 mol.
The moles of nitric acid remaining will be 0.2381 mol - 0.734 mol = -0.496 mol.
However, since moles cannot be negative, the moles of nitric acid remaining will be 0 mol.
Therefore, the mass of nitric acid remaining will be 0 mol * 63.01 g/mol = 0 g.
Based on the calculations, it is clear that nitric acid is the reactant in excess because it is completely consumed in the reaction while there is still magnesium carbonate remaining.