Answer to Question #90330 in General Chemistry for karla

Question #90330
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
1
Expert's answer
2019-05-28T03:17:22-0400

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


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