Question

***Please show ALL work and explain***

If 2.000 g NaHCO3 and 2.0 mL HCl (density = 1.18 g/mL) are reacted to yield 1.100 g NaCl, what is the theoreticaal and percent yield of this reaction?

Answer 1

1)

Molar mass of NaHCO3,

MM = 1*MM(Na) + 1*MM(H) + 1*MM(C) + 3*MM(O)

= 1*22.99 + 1*1.008 + 1*12.01 + 3*16.0

= 84.008 g/mol

mass(NaHCO3)= 2.0 g

use:

number of mol of NaHCO3,

n = mass of NaHCO3/molar mass of NaHCO3

=(2 g)/(84.01 g/mol)

= 2.381*10^-2 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= density * volume

= 1.18 g/mL * 2.0 mL

= 2.36 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(2.36 g)/(36.46 g/mol)

= 6.473*10^-2 mol

Balanced chemical equation is:

NaHCO3 + HCl ---> NaCl + H2CO3

1 mol of NaHCO3 reacts with 1 mol of HCl

for 2.381*10^-2 mol of NaHCO3, 2.381*10^-2 mol of HCl is required

But we have 6.473*10^-2 mol of HCl

so, NaHCO3 is limiting reagent

we will use NaHCO3 in further calculation

Molar mass of NaCl,

MM = 1*MM(Na) + 1*MM(Cl)

= 1*22.99 + 1*35.45

= 58.44 g/mol

According to balanced equation

mol of NaCl formed = (1/1)* moles of NaHCO3

= (1/1)*2.381*10^-2

= 2.381*10^-2 mol

use:

mass of NaCl = number of mol * molar mass

= 2.381*10^-2*58.44

= 1.391 g

Answer: 1.391 g

2)

% yield = actual mass*100/theoretical mass

= 1.1*100/1.391

= 79.06 %

Answer: 79.06 %