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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

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BuyFindarrow_forward

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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Natural waters often contain relatively high levels of calcium ion, Ca 2 + , and hydrogen carbonate ion (bicarbonate), HCO 3 , from the leaching of minerals into the water. When such water is used commercially or in the home, heating of the water leads to the formation of solid calcium carbonate, CaCO 3 , which forms a deposit (“scale”) on the interior of boilers, pipes, and other plumbing fixtures.

:math> Ca ( HCO 3 ) 2 ( a q ) CaCO 3 ( s ) + CO 2 ( g ) + H 2 O ( l )

a sample of well water contains 2.0 × 10 3 mg of Ca ( HCO 3 ) 2 per milliliter, what mass of CaCO 3 scale would 1.0 mL of this water he capable of depositing?

Interpretation Introduction

Interpretation:

To determine the mass of calcium carbonate per 1.0 ml.

Concept Introduction:

A balanced chemical equation is an equation that contains same number of atoms as well as of each element of reactants and products of reaction.

The limiting reactant in a particular reaction has due to following properties:

  1. Limiting reactant completely reacted in a particular reaction.
  2. Limiting reactant determines the amount of the product in mole.
Explanation

Theoretical yield is also known as calculated yield. Theoretical yield is the amount of product which is theatrically calculates by the amount of limiting agent. It is the maximum amount of product which is formed in any reaction.

Other name of actual yield is observed yield.

The percentage yield is calculated by the actual or observed experimental yield divided by the theoretical yield and multiplied by 100.

The percentage yield can be calculates by the use of following expression:

Actual yield or given yield Theoretical yield or calculated yield ×100%   

Theoretical yield is the greatest amount of a product possible from a reaction, when assume that the reaction goes to 100 % competition.

The balance equation for the calcium and oxalate ions is as follows:

Ca(HCO3)2(aq)CaCO3(s)+CO2(g)+H2O(l)

Given:

Amount of Ca(HCO3)2 = 2.0×103 mg per mL

Calculation:

Number of moles of Ca(HCO3)2 is calculated as follows:

Number of moles=mass in gmolarmass=2.0×103 mg×1000g1.00mg162

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