How many moles of AgNO3 are present in 25.0 ml of 0.600 M AgNO3 solution? How many grams of NaOH are required to prepare 0.250 liter of 0.300 M solution? Determine the number of gr... How many moles of AgNO3 are present in 25.0 ml of 0.600 M AgNO3 solution? How many grams of NaOH are required to prepare 0.250 liter of 0.300 M solution? Determine the number of grams of NaHCO3 that one soda mint tablet contains. Read more

Steps to Solve

1. Identify the Required Moles of NaHCO₃ To determine the number of grams of NaHCO₃ in the soda mint tablet, we first need to calculate the moles of HCl that react with NaHCO₃ using the formula for molarity: $$ \text{Moles of HCl} = M \times V $$ Here, ( M ) is the molarity of the HCl solution (0.138 M) and ( V ) is the volume in liters (34.5 mL = 0.0345 L).

2. Calculate Moles of HCl Substituting the values: $$ \text{Moles of HCl} = 0.138 , \text{mol/L} \times 0.0345 , \text{L} = 0.004761 , \text{moles} $$

3. Use Stoichiometry to Find Moles of NaHCO₃ The reaction between NaHCO₃ and HCl is: $$ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 $$ From the balanced equation, the molar ratio of NaHCO₃ to HCl is 1:1. Thus, the moles of NaHCO₃ are equal to the moles of HCl: $$ \text{Moles of NaHCO}_3 = 0.004761 , \text{moles} $$

4. Calculate the Mass of NaHCO₃ To find the mass of NaHCO₃, we use the molar mass of NaHCO₃, which is approximately 84.01 g/mol: $$ \text{Mass of NaHCO}_3 = \text{Moles} \times \text{Molar Mass} = 0.004761 , \text{moles} \times 84.01 , \text{g/mol} = 0.3998 , \text{g} $$

5. Conclusion The number of grams of NaHCO₃ that one tablet contains is approximately 0.400 g.