A 1.230 g
sample of limestone rock is pulverized and then treated with 30.00 mL
of 1.035 M HCl
solution. The excess acid then requires 11.56 mL
of 1.010 M NaOH
for neutralization. Calculate the percent by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl
solution
To calculate the percent by mass of calcium carbonate in the rock, we first need to determine the moles of HCl reacted with the calcium carbonate in the rock, and then use this information to find the moles of calcium carbonate present.
Step 1: Calculate the moles of HCl reacted with the sample of limestone rock.
The concentration of HCl is 1.035 M and the volume used is 30.00 mL (0.03000 L). Using the formula n = M x V, we can calculate the moles of HCl:
n(HCl) = 1.035 mol/L x 0.03000 L = 0.03105 moles
Step 2: Calculate the moles of NaOH used to neutralize the excess HCl.
The volume of NaOH used is 11.56 mL (0.01156 L) and the concentration is 1.010 M. Again, using the formula n = M x V, we can calculate the moles of NaOH:
n(NaOH) = 1.010 mol/L x 0.01156 L = 0.0117076 moles
Step 3: Since NaOH reacts with HCl in a 1:1 ratio, the moles of NaOH used also represents the moles of HCl that were in excess after reacting with the calcium carbonate.
Step 4: Calculate the moles of HCl that reacted with calcium carbonate.
Moles of HCl reacting with the sample = Moles of HCl - Moles of NaOH
= 0.03105 moles - 0.0117076 moles
= 0.0193424 moles
Step 5: Since the reaction between HCl and calcium carbonate is 1:1, the moles of calcium carbonate in the sample is also 0.0193424 moles.
Step 6: Calculate the molar mass of calcium carbonate (CaCO3).
Ca = 40.08 g/mol
C = 12.01 g/mol
O = 16.00 g/mol (x3)
Molar mass of CaCO3 = 40.08 + 12.01 + (16.00 x 3) = 100.09 g/mol
Step 7: Calculate the mass of calcium carbonate in the 1.230 g sample.
Mass of CaCO3 = moles of CaCO3 x molar mass of CaCO3
= 0.0193424 moles x 100.09 g/mol
= 1.935 g
Step 8: Calculate the percent by mass of calcium carbonate in the rock.
Percent CaCO3 = (mass of CaCO3 / mass of sample) x 100%
= (1.935 g / 1.230 g) x 100%
= 157.5%
Therefore, the percent by mass of calcium carbonate in the rock sample is 157.5%.