Question

How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda?
citric acid (C6H8O7) baking soda (NaHCO3)

the equation that demonstrates reaction:
C6H8O7 + 3NaHCO3 → Na3C6H5O7 + 3H2O + 3CO2

Answer 1

Given:

$$c_{C_6H_8O_7}=0.8 M = 0.8 mol/L$$ $$m_{NaHCO_3} = 15 g$$

Solution:

From the reaction equation:

$$3* n_{C_6H_8O_7} = n_{NaHCO_3}$$

Molar mass of baking soda is:

$$M_{NaHCO_3} \approx 23 + 1 + 12 + 3 * 16 = 84 g/mol$$

So,

$$n_{NaHCO_3} = m_{NaHCO_3}/M_{NaHCO_3}$$

And:

$$n_{C_6H_8O_7} = \frac{1}{3}*n_{NaHCO_3}$$

From definition of molarity:

$$V_{C_6H_8O_7} = \frac{n_{C_6H_8O_7}}{c_{C_6H_8O_7}} = \frac{n_{NaHCO_3}}{3* c_{C_6H_8O_7}} = \frac{m_{NaHCO_3}/M_{NaHCO_3} }{3* c_{C_6H_8O_7}}$$

Calculating:

$$V_{C_6H_8O_7} = \frac{15/84 }{3* 0.8}= \frac{5}{84* 0.8} \approx 0.0744 L = 74.4 mL$$

Answer: 74.4 mL