Respuesta :
Answer:
31.9724 mL
Explanation:
The given antacid, Tumsr contains 40% of [tex]CaCO_3[/tex] by mass.
Given mass of Tusmr = 400 mg
Mass of [tex]CaCO_3[/tex] = [tex]\frac {40}{100}\times 400\ mg=160\ mg[/tex]
Also, 1 mg = 0.001 g
Mass of [tex]CaCO_3[/tex] in Tusmr = 0.16 g
Molar mass of [tex]CaCO_3[/tex] = 100.0869 g/mol
Moles = Mass / Molar mass = 0.16 g / 100.0869 g/mol = 15.9862 × 10⁻⁴ moles
Considering the reaction:
[tex]CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O[/tex]
1 mole of [tex]CaCO_3[/tex] react with 2 moles of HCl
15.9862 × 10⁻⁴ moles of [tex]CaCO_3[/tex] react with 2*15.9862 × 10⁻⁴ moles of HCl
Moles of HCl = 31.9724 × 10⁻⁴ moles
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 = 0.100 M
Let, Volume = x mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = x ×10⁻³ L
Moles of HCl = 31.9724 × 10⁻⁴
[tex]31.9724\times 10^{-4}=0.100 \times {x\times 10^{-3}}\ moles[/tex]
Volume = x = 31.9724 mL
The volume, in mL, of 0.100 M HCl that can be neutralized by 400 mg Tumsr will be 32 mL
Stoichiometric calculation
Tumsr contains 40.0 percent CaCO3. Thus, a 400 mg of Tumsr will contain
400 x 40/100 = 160 mg CaCO3
From the equation of the reaction:
CaCO3 + 2HCl ------------> CaCl2 + H2O + CO2
Mole ratio of CaCO3 and HCl = 1:2
Mole of 160 mg CaCO3 = 0.16g/100 = 0.0016 moles
Equivalent mole of HCl = 0.0016 x 2 = 0.0032 moles
Volume of 0.0032 mole, 0.100 M HCl = 0.0032/0.1
= 0.032 L or 32 mL
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