How many moles of BaCl2 are formed in the neutralization of 393 mL of 0.171 M Ba(OH)2 with aqueous HCl?
回答 (1)
2018-03-12 11:09 am
Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O
Mole ratio Ba(OH)₂ : BaCl₂ = 1 : 1
No. of moles of Ba(OH)₂ reacted = (0.171 mol/L) × (393/1000 L) = 0.0672 mol
No. of moles of BaCl₂ formed = 0.0672 mol
Mole ratio Ba(OH)₂ : BaCl₂ = 1 : 1
No. of moles of Ba(OH)₂ reacted = (0.171 mol/L) × (393/1000 L) = 0.0672 mol
No. of moles of BaCl₂ formed = 0.0672 mol
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