Question

Based upon the mass of baking soda (NaHCO3) and using an excess of HCl in this experiment, you will 1) determine the mass of CO2 produced (actual yield); 2) calculate a theoretical yield for CO2; and 3) calculate a percent yield for CO2.

Answer 1

Equation of reaction

NaHCO3 +  HCl  --------->  NaCl  +  H2O  +  CO2(g)

1)The mass(actual yield) of CO2 can be gotten by isolating it from other products and getting its mass.

Assume 1.0g of CO2 was gotten as a product of the experiment

2) For the theoretical yield

The mass of NaHCO3 was not stated, but for the purpose of this solution, I would assume 2g of NaHCO3 was used for the experiment. You can substitute any parameter by following the steps I follow

Number of mole of NaHCO3 = Mass of NaHCO3/ Molar Mass of NaHCO3

Number of mole of NaHCO3 = 2/84.01 = 0.0238 moles

1 mole of NaHCO3 yielded 1 mole of CO2

0.0238 moles of NaHCO3 will yield 0.0238 moles of CO2

Mass of CO2 = Number of moles * Molar Mass

Mass of CO2 = 0.0238 * 44 = 1.0472g

Theoretical yield of CO2 = 1.0472 grams

3) Percentage yield of CO2 = Actual yield/Theoretical yield *100%

Percentage yield of CO2 = 1.0/1.0472 *100

Percentage yield of CO2 = 95.49%

Answer 2

Answer:

1) The mass of $CO_{2}$ produced is 1.29g

2) The theoretical yield for $CO_{2}$ is 1.311g

3) The percent yield is 98.4%

Explanation:

Step 1: in an experiment let's allow sodium bicarbonate (baking soda) to react with hydrochloric acid HCl to obtain high yield $CO_{2}$.

$NaHCO_{3}+HCl$ ⇒ $NaCl + H_{2} O + CO_{2}$

from the balanced equation it can be seen that 1 mole of $NaHCO_{3$ produces 1 mole of $CO_{2}$.

Step 2: Let assume 2.5g of $NaHCO_{3$ were used

mole of $NaHCO_{3$ = 2.5g/molecular mass of $NaHCO_{3$

molecular mass of$NaHCO_{3$ = 23+1+12+(16×2) = 84g/mol

converting mass of $NaHCO_{3$ into mole we have:

           $\frac{2.5g}{84gmol^{-1} }$= 0.0298 moles

Step 3: since $NaHCO_{3$ : $CO_{2}$ mole is ratio 1:1, then 0.0298 mole of $NaHCO_{3$ will produce 0.0298 mole of $CO_{2}$

The mass of this amount of $CO_{2}$ = 0.0298 × molecular mass of $CO_{2}$

                = 0.0298 mol × ( 12+(16×2))g/mol = 1.3111g of $CO_{2}$

 therefore; the theoretical yield expected is 1.311g

Step 4: Let assume for the experiment that we obtained 1.29g  of $CO_{2}$ as the actual yield

then, percent yield for $CO_{2}$ = $\frac{Actual yield obtained}{theoretical yield that should have been obtained}$ × 100%

                                             = 1.2g/1.311g × 100% =98.4%