Question

3. Calculate the theoretical value for the number of moles of CO2 that should have been produced in each balloon assuming that 1.45 g of NaHCO3 is present in an antacid tablet. Use stoichiometry (a mole ratio conversion must be present) to find your answers (there should be three: one answer for each balloon). (6 pts)

Answer 1

the theoretical value for the number of moles of
`CO_(2(aq))` is 0.0173 moles

Step-by-step explanation:

The balanced chemical equation for the reaction is represented by:

`H_3C_6H_5O_(7(aq)) + 3NaHCO_(3(aq)) ------>3CO_(2(g))+3H_2O+Na_3C_6H_5O_(7(aq))`

From above equation; we would realize that 3 moles of
`NaHCO_(3(aq))` reacts with
`H_3C_6H_5O_(7(aq))` to produce 3 moles of
`CO_(2(aq))`

However ; the molar mass of
`NaHCO_(3(aq))` = 84 g/mol

mass given for
`NaHCO_(3(aq))` = 1.45 g

therefore , we can calculate the number of moles of
`NaHCO_(3(aq))` by using the expression :

number of moles of
`NaHCO_(3(aq))` =
`(mass \ given)/( molar \ mass)`

number of moles of
`NaHCO_(3(aq))` =
`(1.45)/(84)`

number of moles of
`NaHCO_(3(aq))` = 0.0173 mole

Since the ratio of
`NaHCO_(3(aq))` to
`CO_(2(aq))` is 1:1; that implies that number of moles of
`NaHCO_(3(aq))` is equal to number of moles of
`CO_(2(aq))` produced.

number of moles of
`CO_(2(aq))` =
`(mass \ given)/( molar \ mass)`

0.0173 =
`(mass \ given)/( 44 \ g/mol)`

mass of
`CO_(2(aq))` = 0.0173 × 44

mass of
`CO_(2(aq))` = 0.7612 g

Thus; the theoretical value for the number of moles of
`CO_(2(aq))` is 0.0173 moles