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bonufazy [111]
4 years ago
15

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

Chemistry
1 answer:
valentinak56 [21]4 years ago
5 0
The reaction between LiOH and HCl is;   
LiOH + HCl → LiCl + H₂O

The stoichiometric ratio between LiOH and HCl is 1 : 1
moles of LiOH added = moles of HCl needed to neutralize.

Molar mass of LiOH = 24 g/mol

moles = mass / molar mass
LIOH moles  = 1.65 / 24 = 0.06875 mol
Hence needed HCl moles = 0.06875 mol

Molarity = moles (mol) / Volume (L)
Hence needed HCl volume = moles / molarity                                           
                                            = 0.06875 mol / 0.150 mol/L                                           
                                            = 0.458 L = 458 mL

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802.2 K

Explanation:

To solve this problem we can use the formula:

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Where Tb is the boiling point (in K).

We already know the values of both ΔHvap and ΔSvap, so we calculate Tb:

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2 years ago
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