Explanation The reaction of calcium carbonate with HCl is given as shown below; CaCO 3 ( s ) + 2HCl(aq) → CaCl 2 (aq) + CO 2 (g) + H 2 O(l) Neutralization of HCl by CaCO 3 : Moles of CaCO 3 in 400 mg of CaCO 3 is calculated as shown below; n = m M = 400 × 10 − 3 g 100.1 g ⋅ mol − 1 = 4.00 × 10 − 3 mol From the above equation, it is found that one mole of CaCO 3 neutralizes 2 mol of HCl . Therefore, the amount of moles of HCl neutralized can be calculated can be given as follows; Amount of HCl(mol) = 4.00 × 10 − 3 mol of CaCO 3 × 2 mol of HCl 1 mol of CaCO 3 = 8.00 × 10 − 3 mol Thus the amount of HCl neutralized by CaCO 3 will be 8.00 × 10 − 3 mol . Neutralization of HCl by Mg(OH) 2 : The neutralization reaction between magnesium hydroxide and hydrochloric acid is given as shown below; Mg(OH) 2 ( s ) + 2HCl → MgCl 2 ( aq ) + 2H 2 O ( l ) Moles of Mg(OH) 2 in 150 mg of Mg(OH) 2 is calculated as shown below; n = m M = 150 × 10 − 3 g 58

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM

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ISBN: 9781319399849

Author: ATKINS

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Chapter F, Problem L.11E

Interpretation Introduction

Interpretation:

The mass of HCl that can be neutralized by 400 mg of CaCO3 and 150 mg of Mg(OH)2 has to be calculated.

Concept Introduction:

Mass to mass conversion is done by considering the mass and molar mass of the reactant. From the mass and molar mass of reactant, the number of moles of reactant is calculated. Considering the balanced chemical equation, the number of moles of product formed is calculated using the conversion factor and finally the mass of product obtained is calculated using the molar mass of the product.

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Chapter F, Problem L.10E

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Chapter F Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM

Ch. F - Prob. A.1AST Ch. F - Prob. A.1BST Ch. F - Prob. A.2AST Ch. F - Prob. A.2BST Ch. F - Prob. A.3AST Ch. F - Prob. A.3BST Ch. F - Prob. A.4AST Ch. F - Prob. A.4BST Ch. F - Prob. A.5AST Ch. F - Prob. A.5BST

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