Question #922a3

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Question #922a3

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Answer 1 · 2014-12-04T23:09:40

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

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

$N a O H + H C l \to N a C l + H_{2} O$ $NaOH + HCl -> NaCl + H_2O$

One can see that we have a $1 : 1$ $1:1$ ratio for $N a O H$ $NaOH$ and $H C l$ $HCl$ , the reactants, and for the products, $N a C l$ $NaCl$ and $H_{2} O$ $H_2O$ .

The molar masses of $N a C l$ $NaCl$ and $H_{2} O$ $H_2O$ are $58.5 \frac{g}{m o l}$ $58.5 g/(mol)$ and $18 \frac{g}{m o l}$ $18 g/(mol)$ , respectively. GIven the fact that the final quantities are known, one can determine that

$n_{N a C l} = \frac{87.75 g}{58.5 \frac{g}{m o l}} = 1.5$ $n_(NaCl) = (87.75 g)/(58.5 g/(mol)) = 1.5$ moles of $N a C l$ $NaCl$

and

$n_{H_{2} O} = \frac{27 g}{18 \frac{g}{m o l}} = 1.5$ $n_(H_2O) = (27 g)/(18 g/(mol)) = 1.5$ moles of $H_{2} O$ $H_2O$ produced.

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

$n_{H C l} = \frac{V}{22.4 L} = \frac{33.6}{22.4} = 1.5$ $n_(HCl) = V/(22.4 L) = 33.6/22.4 = 1.5$ moles

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

$n_{N a O H} = 1.5$ $n_(NaOH) = 1.5$ moles

The mass of $N a O H$ $NaOH$ used is

$m_{N a O H} = n_{N a O H} \cdot 40 = 1.5 \cdot 40 = 60 g r a m s$ $m_(NaOH) = n_(NaOH) * 40 = 1.5 * 40 = 60grams$ , where the molar mass of $N a O H$ $NaOH$ is $40 \frac{g}{m o l}$ $40 g/(mol)$ .

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.

$m_{N a O H} + m_{H C l} = m_{N a C l} + m_{H_{2} O}$ $m_(NaOH) + m_(HCl) = m_(NaCl) + m_(H_2O)$

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

Therefore, $114.75 g = 114.75 g$ $114.75g = 114.75g$ .

Answer 2 · 2014-12-04T23:15:33

The mass is 60g

1 mole of gas occupies $22.4 l$ $22.4l$ @ stp

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

$H C l + N a O H \to N a C l + H_{2} O$ $HCl + NaOH rarrNaCl+ H_2O$

So 1 mole $H C l$ $HCl$ reacts with 1 mole $N a O H \to$ $NaOH rarr$ 1 mole $N a C l$ $NaCl$ + 1 mole $H_{2} O$ $H_2O$

$M_{r} N a O H = 40$ $M_(r)NaOH = 40$
$M_{r} H_{2} O = 18$ $M_rH_2O=18$
$M_{r} N a C l = 58.5$ $M_rNaCl=58.5$

So $1.5 m o l H C l$ $1.5molHCl$ reacts with $1.5 m o l N a O H \to 1.5 m o l N a C l + 1.5 m o l H_{2} O$ $1.5molNaOHrarr1.5molNaCl+ 1.5molH_2O$

1.5mol NaOH = $1.5 \times 40 = 60 g$ $1.5xx40=60g$

1.5mol NaCl = $1.5 \times 58.5 = 87.75 g$ $1.5xx58.5=87.75g$

1.5mol $H_{2} O = 1.5 \times 18 = 27.0 g$ $H_2O= 1.5xx18=27.0g$

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

Answer 3 · 2014-12-05T00:36:37

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 ${1.52800kg/m}^{3}$ $"1.52800kg/m"^3"$ , which is equal to $1.52800g/L$ $"1.52800g/L"$ . 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 = $\frac{36.46g/mol}{1.52800g/L}$ $"36.46g/mol"/"1.52800g/L"$ = $23.86L/mol$ $"23.86L/mol"$

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

We are ready to begin.

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

$NaOH$ $"NaOH"$ + $HCl$ $"HCl"$ $\to$ $rarr$ $NaCl$ $"NaCl"$ + $H_{2} O$ $"H"_2"O"$

All of the mole ratios are $1:1$ $"1:1"$ .

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

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

Determine the mass of HCl gas in 1.41mol.

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

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

Total mass of products = $87.75g + 27g = 114.75g$ $"87.75g + 27g = 114.75g"$

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 , $33.6L$ $"33.6L"$ of $HCl$ $"HCl"$ gas would produce $56.00g NaOH$ $"56.00g NaOH"$ , $82.40g NaCl$ $"82.40g NaCl"$ and ${25.41 g H}_{2} O$ $"25.41 g H"_2"O"$ . Refer to the following equations:

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

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

$1.41mol HCl gas$ $"1.41mol HCl gas"$ x $\frac{1mol H2O}{1mol HCl}$ $"1mol H2O"/"1mol HCl"$ x $\frac{18.02g H2O}{1 mol H2O}$ $"18.02g H2O"/"1 mol H2O"$ = ${25.41g H}_{2} O$ $"25.41g H"_2"O"$

Using the molar volume at STP , which is $22.414L/mol$ $"22.414L/mol"$ , that would produce $1.5 mol HCl$ $"1.5 mol HCl"$ , which would produce ${54.69g HCl gas, 60.00g NaOH, 87.66g NaCl (not 87.75g), and 27.03g H}_{2} O$ $"54.69g HCl gas, 60.00g NaOH, 87.66g NaCl (not 87.75g), and 27.03g H"_2"O"$ . In this case, your instructor may be looking for the mass of $NaOH$ $"NaOH"$ as $60.00g$ $"60.00g"$ . $(87.66g + 27.03g) - 54.69g = 60.00g$ $"(87.66g + 27.03g) - 54.69g = 60.00g"$ . Its possible the instructor may have entered NTP in error.

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