A 1.248-g sample of limestone rock is pulverized and then
treate...

Question

A 1.248-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. 
Calculate the percentage by mass of calcium carbonate 
in the rock, assuming that it is the only substance reacting 
with the HCl solution.

Accepted Answer

Hi everyone hearing the question telling us that a 2.591 g sample is dissolved in 50 ml Of 1.045 molar hydrochloric solution. The solution needed 12.39 ml of 1.293 moller potassium hydroxide to neutralize the excess acid. If magnesium carbonate is the only compound that reacts with hydrochloric acid, what is its percentage in the sample? So first we need to balanced reactions and we're going to have magnesium carbonate plus hydrochloric acid, forms magnesium chloride plus water plus carbon dioxide. And now we need to balance this and if we look we have two chlorine on the on the product side and only one on the reactive side. So we're going to add a two in front of our hydrochloric acid. And our next balanced equation is hydrochloric acid plus potassium hydroxide forms potassium chloride plus water. And this one is already balanced. So our total moles of hydrochloric acid minus the excess moles of hydrochloric acid is going to equal the moles of hydrochloric acid reacted. Our grams of magnesium carbonate equals the moles, a magnesium carbonate times our molar mass Our mass percent equals the grams of magnesium carbonite divided by the grams of our sample times 100. So now we have to solve this. So we're going to plug in what we know. So 1.045 moles of hydrochloric acid Times 0.05 l Equals 0. 225 moles of hydrochloric acid, total one . moles a sodium hydroxide times zero point zero 12 leaders Equals 0.01602 malls of hydrochloric acid in excess. So 0. total minus zero point X Equals 0. - three. Knowles of hydrochloric acid reacted. Now we're going to have 0. 3623 moles of hydrochloric acid Times are multiple ratio. So one mole of magnesium carbonate over two moles of hydrochloric acid times .3139 grams of magnesium carbonate, which is its molar mass. Over one mole of magnesium carbonite equals one point 52 73 grams of magnesium carbonate, so one . g a magnesium carbonate Over Our 2.591 g sample times 100 equals .95%. And that is our final answer. Thank you for watching. Bye.

Watch next