How many grams od mg(oh)2 will be needed to neutralize 25 mlod
stomach acid if stomach acid is 0.010 m hcl?

0.0073 g

Explanation:

Considering:

Or,

Given :

For HCl :

Molarity = 0.010 M

Volume = 25 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25×10⁻³ L

Thus, moles of HCl :

Moles of HCl  = 0.00025 moles

Considering the reaction shown below as:-

2 moles of HCl reacts with 1 mole of

So,

1 moles of HCl reacts with 1/2 mole of

Also,

0.00025 moles of HCl reacts with mole of

Mole of = 0.000125 moles

Molar mass of = 58.3197 g/mol

Mass = Moles*Molar mass = = 0.0073 g

We need 0.00729 grams of Mg(OH)2 (or 7.29 miligrams)

Explanation:

Step 1: Data given

Volume of stomach acid (HCl) = 25 mL =0.025 L

Molarity HCl = 0.010 M

Step 2: The balanced equation

Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O

For 1 mol of Mg(OH)2 we need 2 moles of HCl to produce 1 mol of MgCl2 and 2 moles of H2O

Step 3: Calculate moles of HCl

Moles HCl = molarity * volume

Moles HCl = 0.010 M * 0.025 L

Moles HCl = 0.00025 moles

Step 4: Calculate moles of Mg(OH)2

For 1 mol of Mg(OH)2 we need 2 moles of HCl to produce 1 mol of MgCl2 and 2 moles of H2O

For 0.00025 moles of HCl we need 0.00025/2 = 0.000125 moles of Mg(OH)2

Step 5: Calculate mass of Mg(OH)2

Mass of Mg(OH)2 = moles * molar mass

Mass of Mg(OH)2 = 0.000125 moles * 58.32 g/mol

Mass of Mg(OH)2 = 0.00729 grams = 7.29 mg

We need 0.00729 grams of Mg(OH)2 (or 7.29 miligrams)

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

samuel de champlain. in 1608, king henry iv sponsored pierre dugua, sieur de mons and samuel de champlain as founders of the city of quebec with 28 men. this was the second permanent french settlement in the colony of canada.