Question

A student uses 0.10 M HCl to simulate the acid concentration in the stomach. What volume( in ml) of “stomach acid” reacts with a tablet that contains 0.10 g of magnesium hydroxide?

Answer 1

To find the volume of stomach acid that reacts with the tablet containing magnesium hydroxide, you need to calculate the number of moles of magnesium hydroxide and use the 1:2 ratio with HCl. The volume of stomach acid can then be determined using the molarity of the HCl solution.

Step-by-step explanation:

To determine the volume of "stomach acid" (HCl) that reacts with the tablet containing magnesium hydroxide, we first need to calculate the number of moles of Mg(OH)2 in the tablet. The molar mass of Mg(OH)2 is 58.33 g/mol, so the number of moles is:

(0.10 g) / (58.33 g/mol) = 0.00171 mol

Since HCl and Mg(OH)2 react in a 1:2 ratio according to the balanced chemical equation:

HCl + 2Mg(OH)2 → MgCl2 + 2H2O

The moles of HCl needed would be twice the moles of Mg(OH)2:

0.00171 mol HCl x 2 = 0.00342 mol HCl

Finally, we can use the molarity of the HCl solution (0.10 M) to calculate the volume of HCl needed:

Volume (in mL) = (0.00342 mol) / (0.10 mol/L) x 1000 = 34.2 mL

Therefore, a volume of 34.2 mL of "stomach acid" (0.10 M HCl) would react with the tablet containing 0.10 g of magnesium hydroxide.

Answer 2

Given, concentration of HCl = 0.10 M

Mass of Mg(OH)₂ = 0.10 g

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Moles = mass /molar mass

Mole of Mg(OH)₂ = 0.10 g/ 58.3197 g/mol

Mole of Mg(OH)₂ = 1.71

Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l)

According to the balanced equation 1 moles of Mg(OH)₂ reacts with 2 moles of HCl.

Therefore, 1.71 moles of Mg(OH)₂ reacts with (2 x 1.71) or 3.42 moles of HCl

Molarity = moles / volume in L

Volume of HCl = 3.42 moles of HCl / 0.10 M HCl

Volume of HCl = 34.2 L or 34200 ml

34200 ml of “stomach acid” reacts with a tablet that contains 0.10 g of magnesium hydroxide.