A 150-mL sample of a .600-mol/L calcium nitrate solution is mixed with a 1.00L of .500-mol/L NaOH(aq). Calculate the mass of precipitate produced in this reacton.
mols Ca(NO3)2 = M x L = ?
mols NaOH = M x L = ?
.....Ca(NO3)2 + 2NaOH ==> Ca(OH)2 + 2H2O
Convert mols Ca(NO3)2 to mols Ca(OH)2.
Convert mols NaOH to mols Ca(OH)2.
In limiting reagent problems the smaller amount is ALWAYS the correct amount of ppt formed.
g ppt formed = mols x molar mass.
To calculate the mass of the precipitate produced in the reaction, we first need to determine the chemical equation of the reaction between calcium nitrate (Ca(NO3)2) and sodium hydroxide (NaOH).
The balanced chemical equation for the reaction between calcium nitrate and sodium hydroxide is:
Ca(NO3)2 + 2NaOH → Ca(OH)2 + 2NaNO3
From the balanced equation, we can see that one mole of calcium nitrate reacts with two moles of sodium hydroxide to produce one mole of calcium hydroxide and two moles of sodium nitrate.
Now, let's calculate the number of moles of calcium nitrate and sodium hydroxide in the given solutions:
Volume of calcium nitrate solution = 150 mL = 0.150 L
Concentration of calcium nitrate solution = 0.600 mol/L
Number of moles of calcium nitrate = Concentration × Volume
= 0.600 mol/L × 0.150 L
= 0.090 mol
Volume of sodium hydroxide solution = 1.00 L
Concentration of sodium hydroxide solution = 0.500 mol/L
Number of moles of sodium hydroxide = Concentration × Volume
= 0.500 mol/L × 1.00 L
= 0.500 mol
Now, we need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed in the reaction and determines the amount of product that can form. In this case, calcium nitrate and sodium hydroxide react in a 1:2 mole ratio.
The mole ratio between calcium nitrate and sodium hydroxide is 1:2. Therefore, the 0.090 mol of calcium nitrate requires 2 × 0.090 mol of sodium hydroxide for complete reaction, which is 0.180 mol.
Since we have 0.500 mol of sodium hydroxide, which is more than the required amount of 0.180 mol, it is in excess. Therefore, sodium hydroxide is the excess reagent, and calcium nitrate is the limiting reagent.
The molar mass of calcium hydroxide (Ca(OH)2) can be calculated by adding the atomic masses of calcium (Ca), oxygen (O), and hydrogen (H):
Molar mass of Ca(OH)2 = (40.08 g/mol + 2 × 16.00 g/mol + 2 × 1.01 g/mol)
= 74.10 g/mol
Now, we can calculate the mass of calcium hydroxide (precipitate) formed:
Mass of calcium hydroxide (precipitate) = Number of moles of calcium hydroxide × Molar mass of Ca(OH)2
Number of moles of calcium hydroxide = Number of moles of calcium nitrate (since it is the limiting reagent) = 0.090 mol
Mass of calcium hydroxide (precipitate) = 0.090 mol × 74.10 g/mol
= 6.69 g
Therefore, the mass of precipitate produced in the reaction is 6.69 grams.