Question #d2c5a

1 Answer
May 17, 2017

Answer:

Warning! Long Answer. a. The limiting reactant is #"NaCl"#; b. The mass of #"O"_2# required is 319.6 g; c. The mass of #"Na"_2"SO"_4# produced is 2341 g.

Explanation:

This is a limiting reactant problem.

We know that we will need a balanced equation, masses, and moles, so let's gather all the information in one place.

#M_text(r):color(white)(mmmmll)58.44color(white)(mml)64.06color(white)(mmmmm)32.00color(white)(mll)142.04#
#color(white)(mmmmmml)"4NaCl" + color(white)(ll)"2SO"_2 + 2"H"_2"O" + "O"_2 → "2Na"_2"SO"_4 + "4HCl"#
#"Mass/g:"color(white)(mmll)2335color(white)(mmm)1355#
#"Moles:"color(white)(mmm)39.956color(white)(mm)21.152#
#"Divide by:"color(white)(mmll)4color(white)(mmmml)2#
#"Moles rxn:"color(white)(m)9.9889color(white)(mm)10.576#

#"Moles of NaCl" = 2335 color(red)(cancel(color(black)("g NaCl"))) × "1 mol NaCl"/(58.44 color(red)(cancel(color(black)("g NaCl")))) = "39.956 mol NaCl"#

#"Moles of SO"_2 = 1355 color(red)(cancel(color(black)("g SO"_2))) × ("1 mol SO"_2)/(64.06 color(red)(cancel(color(black)("g SO"_2)))) = "21.152 mol SO"_2#

a) Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will produce.

You simply divide the number of moles by the corresponding coefficient in the balanced equation.

I did that for you in the chart above.

#"NaCl""# is the limiting reactant because it gives the fewest "moles of reaction".

b) Mass of #"O"_2# required

#"Mass of O"_2 = "39.956" color(red)(cancel(color(black)("mol NaCl"))) × (1 color(red)(cancel(color(black)("mol O"_2))))/(4color(red)(cancel(color(black)("mol NaCl")))) × ("32.00 g O"_2)/(1 color(red)(cancel(color(black)("mol O"_2)))) = "319.6 g O"_2#

c) Calculate the mass of #"Na"_2"SO"_4#

(i) Calculate the theoretical yield

#"Theoretical yield" = 39.956 color(red)(cancel(color(black)("mol NaCl"))) × (2 color(red)(cancel(color(black)("mol Na"_2"SO"_4))))/(4 color(red)(cancel(color(black)("mol NaCl")))) × ("142.04 g Na"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "2838 g Na"_2"SO"_4#

(ii) Calculate the actual yield

#"Actual yield" = 2388 color(red)(cancel(color(black)("g theoretical"))) × "82.5 g actual"/(100 color(red)(cancel(color(black)("g theoretical")))) = "2341 g actual"#

The mass of of #"Na"_2"SO"_4# produced is 2341 g.