Chapter 1, Problem 1.87QA

Interpretation Introduction

To check

which mixture follows the law of definite proportion.

Answer to Problem 1.87QA

Solution:

Option a. 11.0 g of sodium and 17.0 g of chlorine and Option b. 6.5 g of sodium and 10.0 g of chlorine would react to produce NaCl, with no sodium or chlorine left over.

Explanation of Solution

Here we will check the mole ratio between sodium and chlorine (chloride ions).

The molar mass of Na is 23.00 g/mol and the molar mass of Cl is 35.45 g/mol.

$Moles of Na=1.00 g Na\times \frac{1 mol Na}{23.00 g Na}=0.04347 mol Na$

$Moles of Cl=1.54 g Cl\times \frac{1 mol Cl}{35.45 g Cl}=0.04344 mol Cl$

Now find out mole ratio between sodium ion and chlorine ion.

$\frac{Moles of {Na}^{+}}{Moles of {Cl}^{-}}= \frac{0.04347 mol}{0.04344 mol}\approx 1$

Therefore Na⁺ and Cl^- ion combine in 1:1 mole ratio.

Similarly we check the mole ratio between Na⁺ and Cl^- ion in other mixtures.

Mixture a. 11.0 g of sodium and 17.0 g of chlorine.

$Moles of Na=11.0 g Na\times \frac{1 mol Na}{23.00 g Na}=0.4783 mol Na$

$Moles of Cl=17.0 g Cl\times \frac{1 mol Cl}{35.45 g Cl}=0.4795 mol Cl$

Now find out mole ratio between sodium ion and chlorine ion.

$\frac{Moles of {Na}^{+}}{Moles of {Cl}^{-}}= \frac{0.4783 mol}{0.4795 mol}\approx 1$

Na⁺ and Cl^- ion combine in 1:1 mole ratio. Therefore mixture a. 11.0 g of sodium ion and 17.0 g of chlorine ion produce NaCl with no sodium or chlorine left over.

Mixture b. 6.5 g of sodium and 10.0 g of chlorine.

$Moles of Na=6.5 g Na\times \frac{1 mol Na}{23.00 g Na}=0.2826 mol Na$

$Moles of Cl=10.0 g Cl\times \frac{1 mol Cl}{35.45 g Cl}=0.2821 mol Cl$

Now find out mole ratio between sodium ion and chlorine ion.

$\frac{Moles of {Na}^{+}}{Moles of {Cl}^{-}}= \frac{0.2826 mol}{0.2821 mol}\approx 1$

Na⁺ and Cl^- ion combine in 1:1 mole ratio. Therefore mixture b. 6.5 g of sodium ion and 10.0 g of chlorine ion produce NaCl with no sodium or chlorine left over.

Mixture c. 6.5 g of sodium and 12.0 g of chlorine.

$Moles of Na=6.5 g Na\times \frac{1 mol Na}{23.00 g Na}=0.2826 mol Na$

$Moles of Cl=12.0 g Cl\times \frac{1 mol Cl}{35.45 g Cl}=0.3385 mol Cl$

Now find out mole ratio between sodium ion and chlorine ion.

$\frac{Moles of {Na}^{+}}{Moles of {Cl}^{-}}= \frac{0.2826 mol}{0.3385 mol}=0.8349$

Here, Na⁺ and Cl^- ion do not combine in 1:1 mole ratio. All the sodium ion will get consumed and some mass of chlorine will remain.

Mixture d. 6.5 g of sodium and 8.0 g of chlorine.

$Moles of Na=6.5 g Na\times \frac{1 mol Na}{23.00 g Na}=0.2826 mol Na$

$Moles of Cl=8.0 g Cl\times \frac{1 mol Cl}{35.45 g Cl}=0.2257 mol Cl$

Now find out mole ratio between sodium ion and chlorine ion.

$\frac{Moles of {Na}^{+}}{Moles of {Cl}^{-}}= \frac{0.2826 mol}{0.2257 mol}=1.25$

Here, Na⁺ and Cl^- ion do not combine in 1:1 mole ratio. All the chlorine ion will get consumed and some mass of sodium will remain.

Conclusion:

Option a. 11.0 g of sodium and 17.0 g of chlorine and Option b. 6.5 g of sodium and 10.0 g of chlorine would react to produce NaCl, with no sodium or chlorine left over.