Question

How many grams of caco3 are needed to react with 15.2g of hcl according to the following equation?

Caco3 + 2hcl = cacl2 + co2 + h2o​

Answer 1

To determine the number of grams of CaCO3 needed to react with 15.2g of HCl, we need to use the molar ratios from the balanced equation.

From the balanced equation:
1 mole of CaCO3 reacts with 2 moles of HCl.

Step 1: Convert the mass of HCl to moles.
Using the molar mass of HCl (36.46 g/mol), we can calculate the number of moles:
15.2 g HCl * (1 mol HCl / 36.46 g HCl) = 0.417 mol HCl

Step 2: Use the mole ratio to find the moles of CaCO3.
Since the ratio is 1:2 (CaCO3:HCl), we can calculate the moles of CaCO3:
0.417 mol HCl * (1 mol CaCO3 / 2 mol HCl) = 0.2085 mol CaCO3

Step 3: Convert the moles of CaCO3 to grams.
Using the molar mass of CaCO3 (100.09 g/mol), we can calculate the grams:
0.2085 mol CaCO3 * (100.09 g CaCO3 / 1 mol CaCO3) = 20.86 g CaCO3

Therefore, approximately 20.86 grams of CaCO3 are needed to react with 15.2 grams of HCl.

Answer 2

Answer: 20.6g

Explanation: From the balanced equation, 2 x 36.468 g HCl required 100.09 g of CaCO3 for reaction. Therefore, 15.2 g of HCl required {(100.09 g/2 x 36.468 g)} x 15.2 g = 20.86 g of CaCO3 for reaction. Hence, approx. 20.9 g of CaCO3 are needed to react with 15.2 g of HCl.