CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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Chapter 12, Problem 12.143P
Interpretation Introduction

Interpretation:

The mass of boron nitride that can be prepared from 1.00 metric ton of boric acid is to be calculated.

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass. Also, the amounts of substances in a balanced chemical reaction are stoichiometrically equivalent to each other.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction.

Amount(mol) of excess reactant is reactant left after the formation of maximum amount(mol) of products.

The formula to calculate moles is as follows:

  Amount(mol)=massmolar mass

The ideal gas equation can be expressed as follows:

  PV=nRT        (1)

Here,

P is the pressure.

V is the volume.

T is the temperature.

n is the mole of the gas.

R is the gas constant.

Expert Solution & Answer
Check Mark

Answer to Problem 12.143P

The mass of boron nitride that can be prepared from 1.00 metric ton of boric acid is 2.98×105g.

Explanation of Solution

The reation of step 1 is as follows:

  B(OH)3(s)+3NH3(g)B(NH2)3(s)+3H2O(g)        (2)

The reaction of step 2 is as follows:

  B(NH2)3(s)BN(s)+2NH3(g)        (3)

Add equation (1) and (2) to obtain the overall reaction.

  B(OH)3(s)+3NH3(g)B(NH2)3(s)+3H2O(g)B(NH2)3(s)BN(s)+2NH3(g)_B(OH)3(s)+NH3(g)BN(s)+3H2O(g)

The fractional yield of the overall reaction is calculated as follows:

  Fractional yield=(yield of step1100%)(yield of step2100%)        (4)

Substitute 85.5% for yield of step 1 and 86.8% for yield of step 2 in the equation (4).

  Fractional yield=(85.5%100%)(86.8%100%)=0.74214

The formula to calculate the moles of B(OH)3 is as follows:

  Moles of B(OH)3=[given massof B(OH)3molecular mass of B(OH)3]        (5)

Substitute 1t for the mass of B(OH)3 and 61.83 g/mol for the molar mass of B(OH)3 in the equation (5).

  Moles of B(OH)3=(1t61.83 g/mol)(1000kg1t)(1000g1kg)=1.6173379×104mol

Rearrange the equation (1) to calculate the number of moles of NH3.

  nNH3=PVRT        (6)

Substitute the value 3.07×103kPa for P, 275 K for T, 12.5m3 for V and 0.0821 Latm/Kmol for R in the equation (6).

  nNH3=(3.07×103kPa)(1atm101.325kPa)(12.5m3)(1L103m3)(0.0821 Latm/Kmol)(275 K)=1.6774720×104mol

The molar ratio of reactants is 1:1. Moles of B(OH)3 is less and therefore B(OH)3 is the limiting reagent.

The formula to calculate the mass of BN is as follows:

  Mass ofBN=[moles ofB(OH)3(1molBN1molB(OH)3)(molar mass of BN)(fractional yield of overall reaction)]        (7)

Substitute 1.6173379×104mol for moles of B(OH)3, 0.74214 for fractional yield of overall reaction and 24.82 g/mol for molar mass of BN in the equation (7).

  Mass ofBN=[(1.6173379×104mol)(1molBN1molB(OH)3)(24.82 g/mol)(0.74214)]=2.97912×105g2.98×105g.

Conclusion

The mass of boron nitride that can be prepared from 1.00 metric ton of boric acid is 2.98×105g.

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Chapter 12 Solutions

CHEMISTRY >CUSTOM<

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