How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 15.0 g of hydrochloric acid?

22.7 g of CaCl₂ are produced in the reaction

Explanation:

This is the reaction:

CaCO₃ +  2HCl  →  CaCl₂  + CO₂  +  H₂O

Now, let's determine the limiting reactant.

Let's divide the mass between the molar mass, to find out moles of each reactant.

29 g / 100.08 g/m = 0.289 of carbonate

15 g / 36.45 g/m = 0.411 of acid

1 mol of carbonate must react with 2 moles of acid

0.289 moles of carbonate will react with the double of moles (0.578)

I only have 0.411 of HCl, so the acid is the limiting reactant.

Ratio is 2:1, so I will produce the half of moles, of salt.

0.411 / 2 = 0.205 moles of CaCl₂

Mol . molar mass = mass → 0.205 m . 110.98 g/m = 22.7 g

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error