1. 2.5g limestone chips (CaCO3) were added to a solution containing 1.0 g hydrocholoric acid
a. how much calcium chloride is formed?
b. how much of which reactant is left over?
This is one of the questions. i found limiting reagent a bit confusing.
Thanks
I am going to tell you the steps, and after you follow them and still can't the answer, then I can do a numerical step by step working out for you.
1)Construct a balanced chemical equation based on reaction between calcium carbonate and hydrochloric acid-your products should be calcium chloride, water and carbon dioxide.
2)Calculate the molar masses of CaCO3 and HCl respectively by adding the respectively atomic weights of each atom in each molecule together-atomic weights is available on the periodic table. Remember that n=m/M, so knowing m and M of each reactant, you can calculate n. For example, to calculate the number of moles of calcium carbonate, n=2.5/100.09, follow the same principle to calculate n for HCi
If you calculate it correctly, you will see that HCI is in excess, which means calcium carbonate is the limiting reagent
a) To calculate how much calcium chloride is formed, since calcium carbonate is the limiting reagent and the molar ratios of the reactants and products in this reaction is 1:1, so essentially you know the number of moles of calcium carbonate from your previous calculations(n), since ratio is the same, n=m/M, so you just calculate the molar mass of CaCI2 using data from periodic table, you know n, so to calculate the mass of CaCl2 formed, m=Mn, just times these the number of moles with the molar mass of CaCI2 and you will get the answer
b)Since HCI is calculated to be in excess, it will be the reactant left over, since the reactants react in the ratio of 1:1, so essentially, you just subtract the moles of the limiting agent, CaCO3, from the moles of HCI, and use that excess number of moles in the formula, n=m/M, you know the molar mass of HCI, you know n, so m=Mn, and so by using this formula, you will get the mass of HCI left over-hope this helps
