How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr (aq) + K2CO3 (aq) →2KBr (aq) + CO2 (g) + H2O (l)

Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?

2HBr (aq) + K2CO3 (aq) - --> 2KBr (aq) + CO1 (g) + H2O (l)

Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr (aq) + K2CO3 (aq) → 2KBr (aq) + CO2 (g) + H2O (l)

Number of moles = Volume * Molar Concentration

moles HBr = 0.42L x. 45 M = 0.189 moles HBr

From the stoichiometry of the reaction;

1 mole of K2CO3 reacts with 2 moles of HBr

1 mole = 2 mole

x mole = 0.189

x = 0.189 / 2 = 0.0945 moles

Mass = Number of moles * Molar mass

Mass = 0.0945 * 138.21 = 13.1 g