Q.10.69 How many grams of CaCO3 are requ... [FREE SOLUTION]

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 ,

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

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,

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

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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91影视

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Chemical Analysis

Inorganic Chemistry

Nuclear Chemistry

Chemical Reactions

Chemistry Branches

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?