Chapter 10: Q.10.69 (page 355)

How many grams of $C a C O_{3}$ are required to neutralize 100mL of stomach acid, which is equivalent to 0.0400 M HCL?

Short Answer

Expert verified

$0.20 g$ of $C a C O_{3}$ is required.

Step by step solution

Given Information

the potential of hydrogen (pH) is a scale used to specify the acidity or basicity of a watery solution. Acidic solutions are estimated to have lower pH values than basic or alkaline solutions.

Explanation

The reaction is ,

$2 HCl (a q) + {CaCO}_{3} (s) \to {CaCl}_{2} (a q) + {CO}_{2} (g) + H_{2} O (l)$

Molarity of HCL = $0.040 M$

Molar mass of ${CaCO}_{3}$ = $100.09 g$ for one mole

$2 molHCl = 1 mol {CaCO}_{3}$ from the equation

now,

$100 m L H C L \times \frac{1 LHCl}{1000 mLHCl} \times \frac{0.040 molHCl}{1 LHCl} \times \frac{1 mol {CaCO}_{3}}{2 molHCl} \times \frac{100.09 g {CaCO}_{3}}{1 mol {CaCO}_{3}}$

localid="1653110764064" $= 0.2 g C a C O_{3}$

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How many grams of $C a C O_{3}$ are required to neutralize 100mL of stomach acid, which is equivalent to 0.0400 M HCL?

Short Answer

Step by step solution

Given Information

Explanation

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How many grams of CaCO3 are required to neutralize 100mL of stomach acid, which is equivalent to 0.0400 M HCL?

Short Answer

Step by step solution

Given Information

Explanation

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Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.

How many grams of $C a C O_{3}$ are required to neutralize 100mL of stomach acid, which is equivalent to 0.0400 M HCL?