Question

Suppose 83.2 mL of a 0.134 M solution of Na,SO reacts with 123 mL of a 0.329 M solution of MgCl, to produce MgSO and NaCl as shown in the balanced reaction Na,SO4 (aq)MgCl2 (aq) MgSO4(s)2 NaCl(aq) Determine the limiting reactant for the given reaction. O MgCl2 MgSO O Na,SO4 O NaCl Calculate the mass of MgSO4 that can be produced in the given reaction mass of MgSO4 1.384 g Only 0.335 g of MgSO, are isolated after carrying out the reaction. Calculate the percent yield of MgSO. percent yield: 24.2

Answer 1

1)

volume of Na2SO4, V = 83.2 mL

= 8.32*10^-2 L

use:

number of mol in Na2SO4,

n = Molarity * Volume

= 0.134*8.32*10^-2

= 1.115*10^-2 mol

volume of MgCl2, V = 1.23*10^2 mL

= 0.123 L

use:

number of mol in MgCl2,

n = Molarity * Volume

= 0.329*0.123

= 4.047*10^-2 mol

Balanced chemical equation is:

Na2SO4 + MgCl2 ---> MgSO4 + NaCl

1 mol of Na2SO4 reacts with 1 mol of MgCl2

for 1.115*10^-2 mol of Na2SO4, 1.115*10^-2 mol of MgCl2 is required

But we have 4.047*10^-2 mol of MgCl2

so, Na2SO4 is limiting reagent

Answer: Na2SO4

we will use Na2SO4 in further calculation

2)

Molar mass of MgSO4,

MM = 1*MM(Mg) + 1*MM(S) + 4*MM(O)

= 1*24.31 + 1*32.07 + 4*16.0

= 120.38 g/mol

According to balanced equation

mol of MgSO4 formed = (1/1)* moles of Na2SO4

= (1/1)*1.115*10^-2

= 1.115*10^-2 mol

use:

mass of MgSO4 = number of mol * molar mass

= 1.115*10^-2*1.204*10^2

= 1.342 g

Answer: 1.34 g

3)

% yield = actual mass*100/theoretical mass

= 0.335*100/1.342

= 24.96%

Answer: 25.0 %