Question

I am confused as to how to find the following in a general chemistry lab on stoichometry...

1. Theoretical moles of CO2 lost

2. Theoretical MASS of CO2 lost

3. Actual Mass of CO2 lost

for the experiment....

NaHCO3(s) + HCL (aq) --->NaCl(aq) + CO2 (g) + H2O(l)

The information i gathered from this experiment is:

Mass of Beaker	Mass of beaker + NaHCO3	Mass of NaHCO3	Mass of HCL	Total Mass	Final mass of reagants + beaker	Final mass	Mass loss
140.95 g	141.95 g	1g	25.44	26.44	166.78 g	25.83 g	.61 g

Reagants	Molar Mass	Amount used	Moles
NaHCO3	83.997 g/mol	1 g	.011 mol NaHCO3
HCL	MOLARITY:0.9992 M	25.44 g	.205 mol

This is the information i'm suposed to use for sevreal of these reactions-can someone please walk me through the 3 questions?!?!?!

Answer 1

as Mass of Beaker = 140.95 g

Mass of beaker + NaHCO3 = 141.95 g

Mass of NaHCO3 = 1g

Mass of HCL = 25.44 g  

Total Mass = 26.44 g of reactant

Final mass of reagants + beaker = 166.78 g

total mass of product = 25.83 g

Mass loss = -0.61 as in above reaction on ly CO₂ (g) is allowed to excape there fore theoretical mass of

CO2 = 0.61g

2) as we know that no of moles = given mass / molar mass

n = 0.61 / 44 = 0.1386 moles

3)AS Moles of 0.011 mol NaHCO3 and .205 moles of HCL

therefore as NaHCO3 is in small amount it is limiting reagent

therefore 0.011 mol of NaHCO3 react witrh .205 moles of HCL to produces 0.011 moles of CO2

or actual mass lost = no of moles * molar mass = 0.011 * 44 =0.484g