Stomach acid is essentially 0.10 M HCl. An active ingredient found in a number of popular antacids is CaCO3. Calculate the number of grams of CaCO3 needed to exactly react with 250 mL of stomach acid.
CaCO3(s)+ 2HCl(aq)= CO2(g)+ CaCl2(aq)+ H20(l)
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To calculate the number of grams of CaCO3 needed to react with 250 mL of stomach acid, we need to first convert the volume of stomach acid into moles of HCl. Then, using the balanced chemical equation, we can determine the stoichiometric ratio between HCl and CaCO3 to find the number of moles of CaCO3 required. Finally, we can convert moles of CaCO3 into grams.
Given:
- Concentration of HCl in stomach acid: 0.10 M
- Volume of stomach acid: 250 mL
Step 1: Convert mL of stomach acid to moles of HCl
To convert from mL to L, divide by 1000:
Volume of stomach acid (L) = 250 mL / 1000 = 0.250 L
Use the molarity equation to calculate moles of HCl:
Moles of HCl = Molarity × Volume (L)
Moles of HCl = 0.10 M × 0.250 L = 0.025 moles
Step 2: Determine the stoichiometry between HCl and CaCO3
From the balanced chemical equation:
CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)
The stoichiometric ratio between HCl and CaCO3 is 2:1. This means that for every 2 moles of HCl, 1 mole of CaCO3 is required.
Step 3: Calculate the moles of CaCO3 required
Using the stoichiometric ratio, we can calculate the moles of CaCO3 needed:
Moles of CaCO3 = Moles of HCl ÷ Stoichiometric ratio
Moles of CaCO3 = 0.025 moles ÷ 2 = 0.0125 moles
Step 4: Convert moles of CaCO3 to grams
To convert from moles to grams, we need to multiply by the molar mass of CaCO3.
The molar mass of CaCO3 can be calculated as follows:
- Atomic mass of Ca: 40.08 g/mol
- Atomic mass of C: 12.01 g/mol
- Atomic mass of O (3 oxygens): 16.00 g/mol
Molar mass of CaCO3 = (40.08 g/mol) + (12.01 g/mol) + (16.00 g/mol × 3) = 100.09 g/mol
Now, multiply the moles of CaCO3 by the molar mass to find the grams:
Grams of CaCO3 = Moles of CaCO3 × Molar mass of CaCO3
Grams of CaCO3 = 0.0125 moles × 100.09 g/mol = 1.25 g
Therefore, 1.25 grams of CaCO3 are needed to exactly react with 250 mL of stomach acid.
To calculate the number of grams of CaCO3 needed to react with the stomach acid, we first need to determine the number of moles of HCl present in the acid solution.
Step 1: Convert the volume of the stomach acid from milliliters (mL) to liters (L).
250 mL = 0.250 L
Step 2: Calculate the number of moles of HCl using the given concentration and volume of the stomach acid solution.
Molarity (M) = Moles (mol) / Volume (L)
0.10 M = Moles / 0.250 L
Rearranging the equation for moles gives:
Moles = Molarity × Volume
Moles = 0.10 mol/L × 0.250 L
Moles = 0.025 mol of HCl
Step 3: Identify the stoichiometry ratio between HCl and CaCO3 from the balanced chemical equation.
From the balanced equation, we see that 1 mole of CaCO3 reacts with 2 moles of HCl.
Step 4: Calculate the number of moles of CaCO3 required to react with the stomach acid.
Since the stoichiometry ratio is 1:2, we need double the number of moles of HCl.
Moles of CaCO3 = 2 × Moles of HCl
Moles of CaCO3 = 2 × 0.025 mol
Moles of CaCO3 = 0.050 mol
Step 5: Convert the moles of CaCO3 to grams using the molar mass of CaCO3.
The molar mass of CaCO3 can be calculated by adding up the atomic masses of its elements:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (there are three oxygen atoms)
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + 16.00 g/mol × 3
Molar mass of CaCO3 = 100.09 g/mol
To calculate the mass:
Mass of CaCO3 = Moles × Molar mass
Mass of CaCO3 = 0.050 mol × 100.09 g/mol
Mass of CaCO3 ≈ 5.00 grams
Therefore, approximately 5.00 grams of CaCO3 is needed to react completely with 250 mL of stomach acid.