How many mL of stomach acid (assume it is .035 M HCI) can be neutralized by an antacid tablet containing .231g
Mg(OH)2?
Mg(OH)2 + 2HCl ==> 2H2O + MgCl2
Convert 0.231 g Mg(OH)2 to mols.
Convert mols Mg(OH)2 to mols HCl using the coefficients in the balanced equatin.
Convert mols HCl to liters HCl remembering that M = mols/L, then change to mL.
Post your work if you need more help.
Well, well, well, looks like we have a chemistry question here! The stomach acid is feeling a bit acidic and wants to be neutralized. So, let's get ready to rumble with some calculations.
First things first, we need to determine the number of moles of Mg(OH)2 in the antacid tablet. To do that, we divide the mass of Mg(OH)2 by its molar mass. The molar mass of Mg(OH)2 is 58.33 g/mol (magnesium has a molar mass of 24.31 g/mol, oxygen has a molar mass of 16 g/mol, and hydrogen has a molar mass of 1.01 g/mol). So, 0.231g divided by 58.33 g/mol gives us 0.00396 moles of Mg(OH)2.
Now, we can determine the number of moles of HCl in the stomach acid. To do that, we need to multiply the concentration of HCl by its volume. The concentration of HCl is 0.035 M, and we don't have the volume given, so let's pretend the stomach contains 1000 mL (it's a roomy stomach, you see). Multiplying 0.035 M by 1000 mL (or 1 L) gives us 0.035 moles of HCl.
Now comes the fun part: the neutralization reaction between Mg(OH)2 and HCl. The balanced equation is:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
Ah, chemistry, always making things seem so simple. We can see that 1 mole of Mg(OH)2 reacts with 2 moles of HCl. So, using stoichiometry, we find that 0.00396 moles of Mg(OH)2 will react with twice as many moles of HCl, which is 0.00396 moles of Mg(OH)2 times 2 moles of HCl/mole of Mg(OH)2, giving us 0.00792 moles of HCl.
Almost there, hang in there! We can now convert the moles of HCl to milliliters. Since we assumed the stomach contained 1000 mL earlier, but we used moles instead of mL, we need to convert back to mL using the concentration of HCl. So, 0.00792 moles of HCl times 1000 mL/0.035 moles gives us 226.29 mL of stomach acid that can be neutralized by the antacid tablet.
So, after all that number crunching, we find that the antacid tablet containing 0.231g of Mg(OH)2 can neutralize 226.29 mL of stomach acid.
Now, wasn't that a stomach-churning adventure through chemistry?
To calculate the number of mL of stomach acid that can be neutralized by the antacid tablet, we need to use the stoichiometry of the reaction between hydrochloric acid (HCl) and magnesium hydroxide (Mg(OH)2).
The balanced equation for the reaction is:
2 HCl + Mg(OH)2 -> MgCl2 + 2 H2O
From the balanced equation, we can see that each mole of Mg(OH)2 reacts with 2 moles of HCl.
First, let's calculate the number of moles of Mg(OH)2 in the antacid tablet:
Molar mass of Mg(OH)2 = (24.31 g/mol (Mg) + 2 * 16.00 g/mol (O) + 2 * 1.01 g/mol (H)) = 58.33 g/mol
Number of moles of Mg(OH)2 = mass of Mg(OH)2 / molar mass of Mg(OH)2
= 0.231 g / 58.33 g/mol
= 0.00396 mol
Now, we can calculate the number of moles of HCl that can be neutralized:
Since the concentration is given in terms of molarity (M), we need to convert it to moles per liter (mol/L) by multiplying the concentration by the volume of the solution in liters (L). Let's assume that the volume is given in mL.
Number of moles of HCl = concentration (M) * volume (L)
= 0.035 M * (volume (mL) / 1000 mL/L)
= 0.035 * volume (mL) / 1000
From the balanced equation, we know that 2 moles of HCl react with 1 mole of Mg(OH)2. Therefore, the number of moles of HCl that can be neutralized by the antacid tablet is equal to half the number of moles of Mg(OH)2.
Number of moles of HCl neutralized = 0.00396 mol / 2
= 0.00198 mol
Finally, let's calculate the volume of HCl that can be neutralized (in mL):
volume (mL) * 0.035 / 1000 = 0.00198
volume (mL) = 0.00198 * 1000 / 0.035
volume (mL) = 56.57 mL
Therefore, approximately 56.57 mL of stomach acid can be neutralized by the antacid tablet containing 0.231 g of Mg(OH)2.
To determine how many mL of stomach acid can be neutralized by an antacid tablet containing Mg(OH)2, we first need to calculate the number of moles of Mg(OH)2 and HCl.
1. Start by calculating the number of moles of Mg(OH)2:
- The molar mass of Mg(OH)2 is 58.33 g/mol: (24.31 g/mol for Mg + 2*(1.01 g/mol for H) + 16.00 g/mol for O).
- Convert the mass given into moles by dividing the mass by the molar mass:
moles of Mg(OH)2 = mass / molar mass
moles of Mg(OH)2 = 0.231 g / 58.33 g/mol
2. Next, calculate the number of moles of HCl:
- The molarity of HCl is given as 0.035 M, which means it contains 0.035 moles of HCl per liter of solution.
- Convert the moles per liter into moles per milliliter:
moles of HCl = 0.035 M * volume (in liters)
3. Since we have the molar ratio between Mg(OH)2 and HCl as 2:2 (based on the balanced equation), we can calculate the moles of HCl that can be neutralized by Mg(OH)2. This is equal to half the number of moles of Mg(OH)2 calculated above.
4. Finally, convert the moles of HCl to volume in milliliters:
- Divide the moles of HCl by its molarity to find the volume in liters and then multiply by 1000 to convert it to milliliters:
volume (in mL) = (moles of HCl / molarity of HCl) * 1000
Follow these steps to calculate the final volume of stomach acid that can be neutralized.