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nekit [7.7K]
3 years ago
15

Hydrobromic acid solution of unknown concentration is titrated with a 0.500M LiOH solution.

Chemistry
1 answer:
hram777 [196]3 years ago
5 0

Answer:

1.00 M

Explanation:

Step 1: Write the balanced equation

HBr + LiOH ⇒ LiBr + H₂O

Step 2: Calculate the reacting moles of lithium hydroxide

40.00 mL of 0.500 M solution react. The reacting moles of LiOH are:

40.00 \times 10^{-3} L \times \frac{0.500mol}{L} = 0.0200 mol

Step 3: Calculate the reacting moles of hydrobromic acid

The molar ratio of HBr to LiOH is 1:1. The reacting moles of hydrobromic acid are 1/1 × 0.0200 mol = 0.0200 mol.

Step 4: Calculate the molarity of hydrobromic acid

0.0200 moles of HBr are in 20.00 mL of the solution. The molarity of the hydrobromic acid solution is:

M= \frac{0.0200 mol}{20.00 \times 10^{-3} L } =1.00 M

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<h2>1.25 g of CO_2 would be produced from the complete reaction of 25 mL of 0.833 mol/L HC_3H_3O_2 with excess NaHCO_3 </h2>

Explanation:

To calculate the number of moles for given molarity, we use the equation:

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0.833M=\frac{\text{Moles of} HC_3H_3O_2\times 1000}{25ml}\\\\\text{Moles of} HC_3H_3O_2 =\frac{0.833mol/L\times 25}{1000}=0.0208mol

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According to stoichiometry:

1 mole of HC_2H_3O_2 will give = 1 mole of CO_2

0.0208 moles of HC_2H_3O_2 will give =\frac{1}{1}\times 0.0208=0.0208 moles of CO_2

Mass of HC_2H_3O_2=moles\times {\text {molar mass}}=0.0208\times 60g/mol=1.25g

Thus 1.25 g of CO_2 would be produced from the complete reaction of 25 mL of 0.833 mol/L HC_3H_3O_2 with excess NaHCO_3

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