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leonid [27]
4 years ago
6

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

Chemistry
1 answer:
CaHeK987 [17]4 years ago
4 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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Explanation:

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Rate Law: k[HI]^2


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1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

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t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

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C.

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There are 2 tables that have different values for the CO2 - the tables in answer choices C and D.

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svetoff [14.1K]

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