How much Ca(NO3)2 should be weighed out to have 0.650?
To have 0.650 what of what. Do you want 0.650 g Ca?
107g
To determine how much Ca(NO3)2 should be weighed out to have 0.650 g of Ca, we first need to calculate the molar mass of Ca(NO3)2 and then use stoichiometry to find the appropriate amount.
The molar mass of Ca(NO3)2 can be calculated by adding up the individual atomic masses of its constituent elements:
Ca: atomic mass = 40.08 g/mol
N: atomic mass = 14.01 g/mol
O: atomic mass = 16.00 g/mol (here, we have three oxygen atoms, so the total mass contribution is 3 x 16.00 g/mol)
Molar mass of Ca(NO3)2 = (40.08 g/mol) + (2 x 14.01 g/mol) + (3 x 16.00 g/mol) = 164.08 g/mol
To determine how much Ca(NO3)2 is needed for 0.650 g of Ca, we need to use the mole ratio between Ca and Ca(NO3)2. From the chemical formula, we know that 1 mole of Ca(NO3)2 contains 1 mole of calcium (Ca).
Using the equation:
moles = mass / molar mass
we can calculate the number of moles of Ca:
moles of Ca = 0.650 g / 40.08 g/mol
moles of Ca ≈ 0.0162 mol
Since the mole ratio between Ca and Ca(NO3)2 is 1:1, we need the same number of moles of Ca(NO3)2 as moles of Ca.
Now, we can calculate the mass of Ca(NO3)2 needed:
mass of Ca(NO3)2 = moles of Ca(NO3)2 x molar mass of Ca(NO3)2
mass of Ca(NO3)2 = 0.0162 mol x 164.08 g/mol
mass of Ca(NO3)2 ≈ 2.65 g
Therefore, approximately 2.65 grams of Ca(NO3)2 should be weighed out to have 0.650 g of Ca.