Respuesta :
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:
0.0073 g
Explanation:
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Or,
[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]
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 :
[tex]Moles=0.010 \times {25\times 10^{-3}}\ moles[/tex]
Moles of HCl = 0.00025 moles
Considering the reaction shown below as:-
[tex]2HCl+Mg(OH)_2\rightarrow MgCl_2+2H_2O[/tex]
2 moles of HCl reacts with 1 mole of [tex]Mg(OH)_2[/tex]
So,
1 moles of HCl reacts with 1/2 mole of [tex]Mg(OH)_2[/tex]
Also,
0.00025 moles of HCl reacts with [tex]\frac{1}{2}\times 0.00025[/tex] mole of [tex]Mg(OH)_2[/tex]
Mole of [tex]Mg(OH)_2[/tex] = 0.000125 moles
Molar mass of [tex]Mg(OH)_2[/tex] = 58.3197 g/mol
Mass = Moles*Molar mass = [tex]0.000125\times 58.3197\ g[/tex] = 0.0073 g
