Chapter 11, Problem 93A

(a)

Interpretation Introduction

Interpretation:

Limiting reactant is to be calculated if 40 g of SiO₂ and 40 g of HF are reacted.

Concept introduction:

1 mol is defined as $6.02214076\times {10}^{23}$ particles of any substance.

Molecular mass is defined as the mass of one mole of substance.

The formula to calculate moles of substance

$\text{Moles=}\frac{\begin{array}{ccc}\text{Mass}& \text{of}& \text{substance}\end{array}}{\begin{array}{cccc}\text{Molar}& \text{mass}& \text{of}& \text{substance}\end{array}}$

Answer to Problem 93A

HFact as the limiting reagent.

Explanation of Solution

The given compound is SiO₂.

The formula to calculate formula mass is,

$\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}$

Where,

• $N_1$ represents the number of atoms of first element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of first element.
• $N_2$ represents the number of atoms of second element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of second element.

The atomic mass of silicon is 28.086 g / mol.

The atomic mass of oxygen is $15.99\text{g}/\text{mol}$.

Substitute the value of atomic mass in the above formula.

$\begin{array}{c}\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}\\ =\left(1\times 28.086\right)+\left(2\times 15.99\right)\\ =60.066\text{g/mol}\end{array}$

Molecular mass of SiO₂ is 60.066 g/mol.

The given compound is HF.

The formula to calculate formula mass is,

$\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}$

Where,

• $N_1$ represents the number of atoms of first element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of first element.
• $N_2$ represents the number of atoms of second element in chemical formula.
• ${\text{x}}_1$ represents the atomic mass of second element.

The atomic mass of fluorine is 18.99 g / mol.

The atomic mass of hydrogen is $1.00\text{g}/\text{mol}$.

Substitute the value of atomic mass in the above formula.

$\begin{array}{c}\text{Formula}\text{mass}=\left(N_1\times x_1\right)+\left(N_2\times x_2\right)+\mathrm{...}\\ =\left(1\times 1\right)+\left(1\times 18.99\right)\\ =19.99\text{g/mol}\end{array}$

Molecular mass of HF is 19.99 g/mol.

$\begin{array}{c}numberofmolesofHF=\frac{MassofHF}{MolarmassofHF}\\ =\frac{40g}{19.99g/mol}\\ =2moles\end{array}$

$\begin{array}{c}numberofmolesofSiO2=\frac{MassofSiO2}{MolarmassofSiO2}\\ =\frac{40g}{60.066g/mol}\\ =0.666moles\end{array}$

SiO₂ react with HF

$SiO2(s)+6HF(aq)\to H2SiF6(aq)+2H2O(l)$

From the stoichiometry of reaction,

For SiO_2,

$\begin{array}{l}1molesofSiO2produce=1molesofH2SiF6\\ 0.666molesofSiO2produce=1\times 0.666molesofH2SiF6\\ 0.666molesofSiO2produce=0.666molesofH2SiF6\end{array}$

For HF

$\begin{array}{c}\text{6}\text{moles}\text{of}\text{HF}\text{produce}\text{=}\text{1}\text{moles}\text{of}\text{H2SiF6}\\ \text{1}\text{moles}\text{of}\text{HF}\text{produce}\text{=}\frac{\text{1}}{\text{6}}\text{moles}\text{of}\text{H2SiF6}\\ \text{2×1}\text{moles}\text{of}\text{HF}\text{produce}\text{=}\text{2×}\frac{\text{1}}{\text{6}}\text{moles}\text{of}\text{H2SiF6}\\ \text{2}\text{moles}\text{of}\text{HF}\text{produce}\text{=}\text{0}\text{.333}\text{moles}\text{of}\text{H2SiF6}\end{array}$

The reactant which produce least amount of product is limiting reagent. So, HFact as the limiting reagent.

(b)

Interpretation Introduction

Interpretation:

The mass of excess reactant is to be calculated

Concept introduction:

The reactant which left over after the reaction is excess reactant.

Answer to Problem 93A

20 g of SiO₂ left after completion of reaction.

Explanation of Solution

Reaction will produce H₂SiF₆ according to limiting reagent

Moles of H₂SiF₆was calculated above.

Number of moles of H₂SiF₆produced by limiting reagent is the actual amount of H₂SiF₆

formed and it is equal to 0.333 moles.

0.333 moles of H₂SiF₆ was produced from 0.333 moles of SiO₂ So, remaining SiO₂ = moles of SiO₂ present − moles of SiO₂ consumed

= 0.666 − 0.333 = 0.333

$\begin{array}{c}Massofsubs\tan ce=moles\times molarmassofsubs\tan ce\\ =0.333\times 60.066\\ =20g\end{array}$

c)

Interpretation Introduction

Interpretation : The theoretical yield needs to be calculated.

Concept introduction : Theoretical yield is the maximum amount of product formed.

Answer to Problem 93A

Theoretical yield is 47.98 g of H₂SiF₆

Explanation of Solution

Step 1: Write balance chemical equation

$SiO2(s)+6HF(aq)\to H2SiF6(aq)+2H2O(l)$

Step 2: convert mass of reactant to moles

Molecular mass of HF is 19.99 g/mol.

$\begin{array}{c}numberofmolesofHF=\frac{MassofHF}{MolarmassofHF}\\ =\frac{40g}{19.99g/mol}\\ =2moles\end{array}$

Step 3: Convert moles of reactant to moles of product using mole ratio

$\begin{array}{c}6molesofHFproduce=1molesofH2SiF6\\ 1molesofHFproduce=\frac{1}{6}molesofH2SiF6\\ 2\times 1molesofHFproduce=2\times \frac{1}{6}molesofH2SiF6\\ 2molesofHFproduce=0.333molesofH2SiF6\end{array}$

Step 4: Convert moles of product to mass (theoretical yield)

$\begin{array}{c}Molar\text{}massofH2SiF6=144.092g/mole\\ Mass=Molarmass\times Moles\\ MassofH2SiF6(Theoreticalyield)=144.092\times 0.333=47.98g\end{array}$

d)

Interpretation Introduction

Interpretation:

If 45.8g of H₂SiF₆ was produce, the percent yield needs to be calculated.

Concept introduction:

The percent yield can be calculated as follows:

$PercentYield=\frac{Actualyield}{Theoreticalyield}\times 100$

Answer to Problem 93A

Percent yield = 95.45%

Explanation of Solution

Actual yield = 45.8 g (given)

$\begin{array}{c}PercentYield=\frac{Actualyield}{Theoreticalyield}\times 100\\ =\frac{45.8}{47.98}\times 100=95.45\%\end{array}$