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♥QT♥
Lv 5
♥QT♥ asked in Science & MathematicsChemistry · 9 years ago

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)

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  • harrie2harrie
    Lv 7
    9 years ago
    Favorite Answer

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

    from bal. rxn., 1 mole K2CO3 reacts with 2 moles HBr(1:2)

    0.189 x 1/2 x 138.21 g/mole K2CO3= 13.1 g K2CO3 required to neutralize spill

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