Question

Calculate the mass of carbon gas produced if 6.000g of K2CO3 reacts with 10.00ml of 2.00M HNO3 identify the limiting reagent as indicated a) if the amount of carbon dioxide collected is part a is 0.3542g, calculate the percent yield for the rection b) if the 0.3542g of carbon dioxide were collected at STP (0C and 1 atm), calculate th volumen of gas collected

Answer 1

Number of moles of HNO₃ , n = Molarity x volume in L

                                           = 2.00 M x 0.010 L

                                           = 0.02 moles

Molar mass of K₂CO₃ = (2x39)+12+(3x16) = 138 g/mol

Given mass of K₂CO₃ is , m = 6.000 g

So number of moles of K₂CO₃ is , n = mass/molar mass

                                                     = 6.000 g / 138 (g/mol)

                                                     = 0.043 moles

K₂CO₃ + 2HNO₃ 2KNO₃ + H₂O + CO₂  

According to the balanced equation ,

1 mole of K₂CO₃ reacts with 2 moles of HNO₃   

M mole of K₂CO₃ reacts with 0.02 moles of HNO₃   

M = (0.02x1) / 2

   = 0.01 mol

So 0.043 - 0.01 = 0.033 moles of K₂CO₃ left unreacted so it is the excess reactant

Since all the mass of HNO₃ complelty reacted it is the limiting reactant.

From the balanced equation ,

2 moles of HNO₃ produces 1 mole of CO₂   

0.02 moles of HNO₃ produces N mole of CO₂   

N = (1x0.02) / 2

   = 0.01 moles

So mass of CO₂ , m = number of moles x Molar mass

                               = 0.01 mol x 44 (g/mol)

                               = 0.44 g of CO₂ ---------> Theoretical yield

So percent yield = ( actual yield / Theoretical yield) x 100

                        = (0.3542 / 0.44) x 100

                        = 80.5 %

We know that PV = nRT

Where

T = Temperature = 273 K

P = pressure = 1.0atm

n = No . of moles = 0.01mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas=?

Plug the values we get

V = (nRT)/P

= (0.01x0.0821x273) / 1

= 0.224 L