Problem PageQuestion A chemistry student weighs out of acetic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.

7.92 mL

Explanation:

There is some info missing. I think this is the original question.

A chemistry student weighs out 0.0617g of acetic acid (HCH₃CO₂) into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1300 M NaOH solution.

Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.

First, we will calculate the moles of acetic acid (MW 60.05 g/mol).

0.0617 g × (1 mol/60.05 g) = 1.03 × 10⁻³ mol

Let's consider the neutralization reaction between HCH₃CO₂ and NaOH.

HCH₃CO₂ + NaOH ⇄ NaCH₃CO₂ + H₂O

The molar ratio of HCH₃CO₂ to NaOH is 1:1. The moles of NaOH are 1.03 × 10⁻³ moles.

The volume of 0.1300 M NaOH that contains 1.03 × 10⁻³ mol is:

1.03 × 10⁻³ mol × 1000 mL/0.1300 mol = 7.92 mL

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2