Question

13) How many grams of baking soda, NaHCO3, are needed to react with 169 mL of stomach acid having an HCl concentration of 0.07640 M?The reaction isNaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l).

Answer 1

Given the following reaction:

`\text{NaHCO}_(3(aq))+HCl_((aq))\text{ }\rightarrow NaCl_((aq))+CO_(2(g))+H_2O_((l))`

We want to know how many grams of NaHCO3 will react with 169 mL (0.169 L) HCl with 0.07640 M concentration.

Step 1: Find the moles of HCl

C = n/V

n = CV

n = 0.07640 M x 0.169 L

n = 0.0129 mol

Step 2: Find the moles of NaHCO3 then convert them to mass.

number of moles of NaHCO3 will be determined using the stoichiometry.

The molar ratio between NaHCO3 and HCl is 1:1

Therefore the number of moles of NaHCO3 = 0.0129 mol.

Now we can find the mass:

n = m/M

m = n x M

m = 0.0129 x 84,007 g/mol

m = 1.085 g