Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+CO2(g) How many grams of calcium chloride will be produced when 32.0 g of calcium carbonate is combined with 11.0 g of hydrochloric acid? mass of CaCl2 : g Which reactant is in excess? HCl CaCO3 How many grams of the excess reactant will remain after the reaction is complete? mass of excess reactant: g

Answer 1

Answer: A) Mass CaCl2 is 16,65g

B) Excess Reactant is the CaCO3

C) Mass of R Excess eactant is 16,9g CaCO3

Explanation:

A) In order to calculate the Mass of CaCl2, we first check the Equation is actually Balanced, and then we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant.

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

Moles produce with 32g CaCO3 ⟶ 32.0g CaCO3 x (1mol CaCO3/100.09g CaCO3)x (1mol CaCl2 /1 mol CaCO3) = 0.32 mol CaCl2

Moles produce with 11 g HCl ⟶ 11.0 g HCl x ( 1mol HCL / 36.46g HCl) x (1 mol CaCl2 /2 mol HCl) = 0.15 mol CaCl2

As we can see the amount of CaCl2 formed with the HCl is the lowest one , therefore the limiting reactant is the HCl. Then, we proceed calculating the mass of CaCl2 from the 11g of HCl.

0.15mol CaCl2 x (110.98 g CaCl2 /1mol CaCl2) = 16.65 g CaCl2

B) The Excess Reagent is the one that produces a larger amount of product that we have previously calculated.

In this case is the CaCO3.

C) In order to estimate the mass of excess reagent, we start by calculating how much CaCO3 reacts with the giving HCl:

11.0 g HCl x (1mol HCL/36.46g HCl) x ( 1 mol CaCO3 / 2 mol HCl)x (100.09 g CaCO3/ 1 mol CaCO3) = 15.10 g CaCO3

That means that only 15.10g CaCO3 will react with 11g of HCl however we were giving 32g of CaCO3, thus, 32g - 15.10g = 16.9g CaCO3 left

Answer 2

• A) To get the mass of $CaCl_{2}$, we must first ensure that the Equation is balanced, and then we must determine which reagent is the limiting one by comparing the quantity of product created with each reactant, with the limiting reactant being the one with the lowest number. Moles produce with $32.0g CaCO_{3} x (1mol CaCO_{3} /100.09g CaCO_{3} )x (1mol CaCl_{2} /1 mol CaCO_{3} ) = 0.32 mol CaCl2\\ 11.0 g HCl x ( 1mol HCL / 36.46g HCl) x (1 mol CaCl_{2} /2 mol HCl) = 0.15 mol CaCl_{2}$As can be seen, the amount of $CaCl_{2}$ produced with HCl is the smallest, hence HCl is the limiting reactant. Then, using the 11g of HCl, we calculate the mass of $CaCO_{3}$ .
• 0.15mol $CaCl_{2}$  x (110.98 g
• B) The Excess Reagent is the one that yields a higher amount of product than we determined previously. CaCO3 is the case here.
• C) To calculate the mass of excess reagent, first determine how much CaCO3 reacts with the HCl given:

Thus, that means 15.10g CaCO3 will react with 11g HCl, but we were only supplying 32g CaCO3, therefore $32g - 15.10g = 16.9g CaCO3$ remains.

Thus the answers are: A)  Mass is 16,65g B) Excess Reactant is the CaCO3 C) Mass of R Excess reactant is 16.9g

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