A1.248 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. part a calculate the percent by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the hcl solution.

First, we will find the moles of HCl entered. This will require the use of the formula:

Moles = molarity * volume (in L)
Moles = 1.035 * 0.03
Moles of HCl = 0.0315

Next, we find the amount reacted with NaOH. Note that each mole of NaOH reacts with one mole of HCl (determined from reaction equation)

Moles of NaOH = 1.01 * 0.01156
Moles of NaOH = 0.01168

Moles of HCl reacted with calcium carbonate = 0.0315 - 0.01168
= 0.01982 mole

The reaction equation is:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

So the moles of CaCO₃ were half that of HCl, which means:
0.01982 / 2 = 0.00991

The mass of calcium carbonate is:

Mass = moles * molecular weight
Mass = 0.00991 * 100
Mass = 0.991 grams

Percentage mass = (0.991 / 1.248) * 100

The ore has 79.41% calcium carbonate by mass.

The answer for this question is d. none of these