Rolaids is an over the counter remedy for upset stomachs that functions by reacting with stomach acid. The active ingredients are calcium carbonate (550 mg) and magnesium hydroxide (110 mg). The inactive ingredients are dextrose, flavoring, magnesium stearate,
polyethylene glycol, pregelatinized starch, and sucrose. The stomach can expand to hold up to one liter of food, and when empty, will hold between 45 and 75 mL gastric fluid, which is about 0.1 M HCl (aq). The reactions that occur are:
Mg(OH)2(aq)+HCl(aq)-> H2O(l) +MgCl2(aq) (reacts “first”)
CaCO3(aq)+HCl(aq)-> H2O(l)+ CaCl2(aq) (reacts “second”)
If a person took 2 rolaids while their stomach contained 45 mL gastric fluid, find how much HCl(aq) is consumed and how much, if any, of the Rolaids active ingredients (in mg and identify
which one) remain.
I wonder how one knows that Mg(OH)2 will react first and if so it all of it will react before CaCO3 begins to react. We will assume that is so. You know that mL x M = 45 x 0.1M = 4.5 millimols of acid will be used. How much Mg(OH)2 do you have? That's 110 x 2 x (1 g/1000 mg) x (1/58.32) = approx 0.00378 mols Mg(OH)2. If all of the Mg(OH)2 is consumed it will use0.00378 x (2 mol HCl/1 mol Mg(OH)2) = 0.00756 mols HCl and you have only 0.0045 so HCl is the limiting reagent and you will use all of it and have some Mg(OH)2 and all of the CaCO3 left over. The problem's wording suggests that only ONE reactants remains; I may have missed something but I think both remain. Can you take it from here?
The equations given are not balanced. They are
Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
CaCO3 + 2HCl ==> CaCl2 | H2O + CO2
To determine how much HCl (aq) is consumed and how much of the Rolaids active ingredients remain, we need to calculate the amount of HCl (aq) consumed in the reactions with both magnesium hydroxide (Mg(OH)2) and calcium carbonate (CaCO3).
First, let's calculate the amount of HCl (aq) that reacts with magnesium hydroxide (Mg(OH)2).
1. Calculate the number of moles of Mg(OH)2:
- Since the molecular weight of Mg(OH)2 is 58.32 g/mol (24.31 g/mol for Mg and 2 * 16.00 g/mol for O and H combined), we can find the number of moles using the formula:
moles = mass / molecular weight
moles = 110 mg / 58.32 g/mol = 1.887 × 10^-3 mol (approximately)
2. From the balanced equation, we can see that the ratio between Mg(OH)2 and HCl is 1:2.
Therefore, 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
3. Calculate the amount of HCl (aq) reacted with Mg(OH)2:
moles of HCl (aq) = 2 * moles of Mg(OH)2
moles of HCl (aq) = 2 * 1.887 × 10^-3 mol = 3.774 × 10^-3 mol (approximately)
4. Calculate the volume of HCl (aq) reacted:
The volume of HCl (aq) reacted can be calculated using the molarity (0.1 M) and the number of moles:
volume = moles / molarity
volume = 3.774 × 10^-3 mol / 0.1 M = 3.774 × 10^-2 L or 37.74 mL (approximately)
So, 37.74 mL of HCl (aq) is consumed when 2 Rolaids tablets react with 45 mL of gastric fluid.
Now let's calculate the amount of HCl (aq) that reacts with calcium carbonate (CaCO3).
1. Calculate the number of moles of CaCO3:
- Since the molecular weight of CaCO3 is 100.09 g/mol (40.08 g/mol for Ca, 12.01 g/mol for C, and 3 * 16.00 g/mol for O), we can find the number of moles using the formula:
moles = mass / molecular weight
moles = 550 mg / 100.09 g/mol = 5.496 × 10^-3 mol (approximately)
2. From the balanced equation, we can see that the ratio between CaCO3 and HCl is 1:1.
Therefore, 1 mole of CaCO3 reacts with 1 mole of HCl.
3. Calculate the amount of HCl (aq) reacted with CaCO3:
moles of HCl (aq) = moles of CaCO3
moles of HCl (aq) = 5.496 × 10^-3 mol (approximately)
4. Calculate the volume of HCl (aq) reacted:
volume = moles / molarity
volume = 5.496 × 10^-3 mol / 0.1 M = 5.496 × 10^-2 L or 54.96 mL (approximately)
So, 54.96 mL of HCl (aq) is consumed when 2 Rolaids tablets react with 45 mL of gastric fluid.
To find out how much of the Rolaids active ingredients remain, we need to consider the excess reactants. In this case, we have an excess of magnesium hydroxide (Mg(OH)2) since it reacts "first" before calcium carbonate (CaCO3).
1. Remaining amount of Mg(OH)2:
The amount of Mg(OH)2 remaining can be calculated by subtracting the moles reacted with HCl (aq) from the initial amount of moles:
remaining moles of Mg(OH)2 = initial moles - moles reacted
remaining moles of Mg(OH)2 = 1.887 × 10^-3 mol - 1.887 × 10^-3 mol = 0 mol
Therefore, there is no remaining Mg(OH)2 after the reaction.
2. Remaining amount of CaCO3:
The amount of CaCO3 remaining can be calculated by subtracting the moles reacted with HCl (aq) from the initial amount of moles:
remaining moles of CaCO3 = initial moles - moles reacted
remaining moles of CaCO3 = 5.496 × 10^-3 mol - 5.496 × 10^-3 mol = 0 mol
Again, there is no remaining CaCO3 after the reaction.
So, after reacting with 45 mL of gastric fluid containing 0.1 M HCl (aq), all the active ingredients in the Rolaids tablets are consumed, and there is no remaining magnesium hydroxide (Mg(OH)2) or calcium carbonate (CaCO3).