If 125 g of CaCO3 is mixed and reacted with 125 g of HCl, which reactant is limiting and how many grams of CO2 can be made?

CaCO3 is the limiting reactant

55 g of CO2 is made

Explanation:

First we must put down the reaction equation;

CaCO3(s) + 2HCl(aq) > CaCl2(s) + H2O(l) + CO2(g)

Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles

From the reaction equation;

1 mole of CaCO3 yields 1 mole of CO2

Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2

For HCl;

number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles

From the reaction equation;

2 moles of HCl yields 1 mole of CO2

3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2

Hence CaCO3 is the limiting reactant.

Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%