Question

Question #922a3

Answer 1

The mass of `NaOH` is equal to `60 grams`.

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

`NaOH + HCl -> NaCl + H_2O`

One can see that we have a `1:1` ratio for `NaOH` and `HCl`, the reactants, and for the products, `NaCl` and `H_2O`.

The molar masses of `NaCl` and `H_2O` are `58.5 g/(mol)` and `18 g/(mol)`, respectively. GIven the fact that the final quantities are known, one can determine that

`n_(NaCl) = (87.75 g)/(58.5 g/(mol)) = 1.5` moles of `NaCl`

and

`n_(H_2O) = (27 g)/(18 g/(mol)) = 1.5` moles of `H_2O` produced.

The number of moles of `HCl` can be determined by using

`n_(HCl) = V/(22.4 L) = 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 `NaOH` moles is equal to

`n_(NaOH) = 1.5` moles

The mass of `NaOH` used is

`m_(NaOH) = n_(NaOH) * 40 = 1.5 * 40 = 60grams`, where the molar mass of `NaOH` is `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_(NaOH) + m_(HCl) = m_(NaCl) + m_(H_2O)`

`60g + (1.5 * 36.5)g = 87.75g + 27g` , where `36.5 g/(mol)` is the molar mass of `HCl`.

Therefore, `114.75g = 114.75g`.

Answer 2

The mass is 60g

1 mole of gas occupies `22.4l` @ stp

So no. moles `HCl=(33.6)/(22.4)=1.5`

`HCl + NaOH rarrNaCl+ H_2O`

So 1 mole `HCl` reacts with 1 mole `NaOH rarr` 1 mole `NaCl` + 1 mole `H_2O`

`M_(r)NaOH = 40`
`M_rH_2O=18`
`M_rNaCl=58.5`

So `1.5molHCl` reacts with `1.5molNaOHrarr1.5molNaCl+ 1.5molH_2O`

1.5mol NaOH = `1.5xx40=60g`

1.5mol NaCl =`1.5xx58.5=87.75g`

1.5mol `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

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"`, which is equal to `"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 = `"36.46g/mol"/"1.52800g/L"` = `"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"` + `"HCl"` `rarr` `"NaCl"` + `"H"_2"O"`

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

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

`"33.6L HCl"` x `"1 mol HCl"/"23.86L HCl"` = `"1.41mol HCl(g)"`

Determine the mass of HCl gas in 1.41mol.

`"1.41 mol HCl gas"` x `"36.46g HCl gas"/"1mol 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"`

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"` of `"HCl"` gas would produce `"56.00g NaOH"`, `"82.40g NaCl"` and `"25.41 g H"_2"O"`. Refer to the following equations:

`"1.41mol HCl gas"` x `"1mol NaOH"/"1mol HCl"` x `"39.9997g NaOH"/"1 mol NaOH"` = `"56.40g NaOH"`

`"1.41mol HCl gas"` x `"1mol NaCl"/"1mol HCl"` x `"58.44g NaCl"/"1 mol NaCl"` = `"82.40g NaCl"`

`"1.41mol HCl gas"` x `"1mol H2O"/"1mol HCl"` x `"18.02g H2O"/"1 mol H2O"` = `"25.41g H"_2"O"`

Using the molar volume at STP , which is `"22.414L/mol"`, that would produce `"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"`. In this case, your instructor may be looking for the mass of `"NaOH"` as `"60.00g"`. `"(87.66g + 27.03g) - 54.69g = 60.00g"`. Its possible the instructor may have entered NTP in error.