Question #922a3

3 Answers
Dec 4, 2014

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#.

Dec 4, 2014

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.

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 #"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.