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LUCKY_DIMON [66]
3 years ago
14

To neutralize 1.65g LiOH, how much .150 M HCl would be needed

Chemistry
1 answer:
Vinvika [58]3 years ago
4 0
The molecular weight of LiOH would be 23.95 g/mol, so the amount of LiOH in mol would be: 1.64g/(<span>23.95 g/mol)= 0.069 mol

The reaction of LiOH with HCl would be:
</span><span>HCl + LiOH = H2O + LiCl
The coefficient of LiOH:HCL is 1:1 so you need the same amount of HCl to neutralize LiOH.

HCl= LiOH
volume* 0.15M= </span>0.069 mol
volume= 0.069 mol/ (0.15 mol/ 1000ml)
volume= 459.29 ml
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