Question

How many grams od Mg(OH)2 will be needed to neutralize 25 mlod

Answer 1

Answer:

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)

Answer 2

Answer:

0.0073 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

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 :

$Moles=0.010 \times {25\times 10^{-3}}\ moles$

Moles of HCl  = 0.00025 moles

Considering the reaction shown below as:-

$2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O$

2 moles of HCl reacts with 1 mole of $Mg(OH)_2$

So,

1 moles of HCl reacts with 1/2 mole of $Mg(OH)_2$

Also,

0.00025 moles of HCl reacts with $\frac{1}{2}\times 0.00025$ mole of $Mg(OH)_2$

Mole of $Mg(OH)_2$ = 0.000125 moles

Molar mass of $Mg(OH)_2$ = 58.3197 g/mol

Mass = Moles*Molar mass = $0.000125\times 58.3197\ g$ = 0.0073 g