Question

Trying to solve for row 2I wanted to add column B we are assuming the concentrating of the acetic acid is 0.42M and the volume of acetic acid is 0.050L

Answer 1

Explanations:

The formula for calculating the number of moles of a compound is expressed as:

`\text{Number of moles=}\frac{Mass}{molar\text{ mass}}`

Given that mass of NaHCO3 = 1.0g

The molar mass of NaHCO3 = (1*23) + 1 + 12 + (3*16)

The molar mass of NaHCO3 = 23 + 13 + 48

The molar mass of NaHCO3 = 84g/mol

Get the moles of NaHCO3

`\begin{gathered} \text{Number of moles =}(1.0g)/((84g)/(mol)) \\ \text{Number of moles = }0.0119\text{moles} \\ _{}NumberofmolesofNaHCO_3=\text{1.19}*10^(-2)moles \end{gathered}`

Get the moles of C2H3O2H (acetic acid)

Given the following parameters:

`\begin{gathered} \text{Molar mass of acetic acid = 0.42M} \\ \text{Volume of acetic acid = 0.050L} \end{gathered}`
`\begin{gathered} moles\text{ of acetic acid=Molarity}* volume \\ moles\text{ of acetic acid}=0.42\frac{mol}{\cancel{dm^3}}*0.050\cancel{dm^3} \\ \text{moles of acetic acid=}0.021\text{ moles} \\ moles\text{ of acetic acid = 2.1}*10^(-2)moles \end{gathered}`

The limiting reagent will be the compound with the least number of moles. From the resulting moles, the compound that has the lower number of moles is acetic acid, hence the limiting reagent will be C2H3O2H

The excess reagent will be NaHCO3