How many moles of calcium bicarbonate are needed if 10.0 grams of sodium bicarbonate are used?
To find the number of moles of calcium bicarbonate needed, we need to first determine the molar mass of sodium bicarbonate (NaHCO3) and the molar mass of calcium bicarbonate (Ca(HCO3)2).
The molar mass of sodium bicarbonate (NaHCO3) can be calculated by adding up the atomic masses of its constituent elements: sodium (Na), hydrogen (H), carbon (C), and oxygen (O).
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Molar mass of NaHCO3 = (22.99 g/mol) + (1.01 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol)
= 84.01 g/mol
Next, we can determine the molar mass of calcium bicarbonate (Ca(HCO3)2).
Ca = 40.08 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Molar mass of Ca(HCO3)2 = (40.08 g/mol) + (2 x [1.01 g/mol]) + (2 x [12.01 g/mol]) + (6 x [16.00 g/mol])
= 162.12 g/mol
Now, we can use the molar mass and the given mass of sodium bicarbonate to calculate the number of moles of sodium bicarbonate.
Number of moles of NaHCO3 = (Given mass of NaHCO3) / (Molar mass of NaHCO3)
= 10.0 g / 84.01 g/mol
≈ 0.119 moles
Since calcium bicarbonate (Ca(HCO3)2) has a 1:1 stoichiometric ratio with sodium bicarbonate (NaHCO3), the number of moles of calcium bicarbonate required will be the same as the number of moles of sodium bicarbonate.
Therefore, the number of moles of calcium bicarbonate needed is approximately 0.119 moles.