Astudent 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?

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.

b i think it’s a water wheel turning in a stream

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole