# Question #922a3

Dec 4, 2014

The mass of $N a O H$ is equal to $60 g r a m s$.

First of all, start by writing the balanced chemical equation.

$N a O H + H C l \to N a C l + {H}_{2} O$

One can see that we have a $1 : 1$ ratio for $N a O H$ and $H C l$, the reactants, and for the products, $N a C l$ and ${H}_{2} O$.

The molar masses of $N a C l$ and ${H}_{2} O$ are $58.5 \frac{g}{m o l}$ and $18 \frac{g}{m o l}$, respectively. GIven the fact that the final quantities are known, one can determine that

${n}_{N a C l} = \frac{87.75 g}{58.5 \frac{g}{m o l}} = 1.5$ moles of $N a C l$

and

${n}_{{H}_{2} O} = \frac{27 g}{18 \frac{g}{m o l}} = 1.5$ moles of ${H}_{2} O$ produced.

The number of moles of $H C l$ can be determined by using

${n}_{H C l} = \frac{V}{22.4 L} = \frac{33.6}{22.4} = 1.5$ moles

Therefore, knowing the $1 : 1$ mole ratio between the reactants and the products, we can determine that the number of $N a O H$ moles is equal to

${n}_{N a O H} = 1.5$ moles

The mass of $N a O H$ used is

${m}_{N a O H} = {n}_{N a O H} \cdot 40 = 1.5 \cdot 40 = 60 g r a m s$, where the molar mass of $N a O H$ is $40 \frac{g}{m o l}$.

Now, according to the law of conservation of mass, the total mass of the reactans must be equal to the total mass of the products.

${m}_{N a O H} + {m}_{H C l} = {m}_{N a C l} + {m}_{{H}_{2} O}$

$60 g + \left(1.5 \cdot 36.5\right) g = 87.75 g + 27 g$ , where $36.5 \frac{g}{m o l}$ is the molar mass of $H C l$.

Therefore, $114.75 g = 114.75 g$.

Dec 4, 2014

The mass is 60g

1 mole of gas occupies $22.4 l$ @ stp

So no. moles $H C l = \frac{33.6}{22.4} = 1.5$

$H C l + N a O H \rightarrow N a C l + {H}_{2} O$

So 1 mole $H C l$ reacts with 1 mole $N a O H \rightarrow$ 1 mole $N a C l$ + 1 mole ${H}_{2} O$

${M}_{r} N a O H = 40$
${M}_{r} {H}_{2} O = 18$
${M}_{r} N a C l = 58.5$

So $1.5 m o l H C l$ reacts with $1.5 m o l N a O H \rightarrow 1.5 m o l N a C l + 1.5 m o l {H}_{2} O$

1.5mol NaOH = $1.5 \times 40 = 60 g$

1.5mol NaCl =$1.5 \times 58.5 = 87.75 g$

1.5mol ${H}_{2} O = 1.5 \times 18 = 27.0 g$

nb I used 22.4 for the molar volume which is at stp, not ntp which is 24L.

Dec 5, 2014

At STP (273.15K and 1 atm) the molar volume of a gas is 22.414L/mol. However, the reaction in this question takes place at NTP (293.15K and 1 atm). The molar volume of a gas at 293.15K is not the same as the molar volume at 273.15K.

To calculate the molar volume of a gas at NTP, you must divide its molar mass by its density at NTP. n.wikipedia.org/wiki/Molar_volume

The density of HCl gas at NTP is $\text{1.52800kg/m"^3}$, which is equal to $\text{1.52800g/L}$. http://www.engineeringtoolbox.com/gas-density-d_158.html

The molar mass of HCl is 36.46g/mol.

Molar volume for a gas at NTP = $\text{36.46g/mol"/"1.52800g/L}$ = $\text{23.86L/mol}$

The molar volume of a gas used to solve this problem will be 23.86L/mol.

Write the balanced equation so that you can determine the mole ratios of the reactants and products. .

$\text{NaOH}$ + $\text{HCl}$ $\rightarrow$ $\text{NaCl}$ + $\text{H"_2"O}$

All of the mole ratios are $\text{1:1}$.

Determine the number of moles of HCl gas used in the reaction.

$\text{33.6L HCl}$ x $\text{1 mol HCl"/"23.86L HCl}$ = $\text{1.41mol HCl(g)}$

Determine the mass of HCl gas in 1.41mol.

$\text{1.41 mol HCl gas}$ x $\text{36.46g HCl gas"/"1mol HCl gas}$ = $\text{51.41g HCl gas}$

Now lets go back to the given equation with the masses of the products given.

Total mass of products = $\text{87.75g + 27g = 114.75g}$

Assuming the validity of the law of conservation of mass, the total mass of the reactants must equal the total mass of the products, I believe this is what your teacher is looking for:

The mass of NaOH = the mass of the products - the mass of HCl = 114.75g - 51.41g = 63.33g NaOH. However, this is inaccurate.

By using stoichiometry, at NTP , $\text{33.6L}$ of $\text{HCl}$ gas would produce $\text{56.00g NaOH}$, $\text{82.40g NaCl}$ and $\text{25.41 g H"_2"O}$. Refer to the following equations:

$\text{1.41mol HCl gas}$ x $\text{1mol NaOH"/"1mol HCl}$ x $\text{39.9997g NaOH"/"1 mol NaOH}$ = $\text{56.40g NaOH}$

$\text{1.41mol HCl gas}$ x $\text{1mol NaCl"/"1mol HCl}$ x $\text{58.44g NaCl"/"1 mol NaCl}$ = $\text{82.40g NaCl}$

$\text{1.41mol HCl gas}$ x $\text{1mol H2O"/"1mol HCl}$ x $\text{18.02g H2O"/"1 mol H2O}$ = $\text{25.41g H"_2"O}$

Using the molar volume at STP , which is $\text{22.414L/mol}$, that would produce $\text{1.5 mol HCl}$, which would produce $\text{54.69g HCl gas, 60.00g NaOH, 87.66g NaCl (not 87.75g), and 27.03g H"_2"O}$. In this case, your instructor may be looking for the mass of $\text{NaOH}$ as $\text{60.00g}$. $\text{(87.66g + 27.03g) - 54.69g = 60.00g}$. Its possible the instructor may have entered NTP in error.