Question

What is the minimum amount of 6.0 M H2SO4H2SO4 necessary to produce 25.0 g of H2(g)H2(g) according to the reaction between aluminum and sulfuric acid

Answer 1

2.083 Liters of 6.0 M solution sulfuric acid is required. This solved using molecular calculations and Titration.

Solution: $2Al(s)+3H_2SO_4(aq) = Al_2(SO_4)_3(aq)+3H_2(g)$

Moles of hydrogen gas =  $\frac{25}{2} = 12.5 mol$

Then 12.5 moles of hydrogen will be obtained from Moles of Sulfuric acid = 12.5 mol

Molarity of the sulfuric acid solution = 6.0 M = 6 mol/ l

6M = $\frac{12.5 mole}{V}$

where V is the volume needed

$V = \frac{12.5}{6}$

V = 2.083 l

What is Titration?

• Titration, commonly referred to as titrimetry, is a typical quantitative chemical analysis method used in laboratories to ascertain the unidentified quantity of an analyte .
• Titration is frequently referred to as volumetric analysis because it relies heavily on volume measurements. The titrant or titrator is a reagent that is prepared as a standard solution.
• To determine concentration, a solution of the analyte or titrand reacts with a known concentration and volume of the titrant. The titration volume is the amount of titrant that has responded.
• Titrations come in a variety of forms with various protocols and objectives. Redox and acid-base titrations are the two most typical types of qualitative titrations.

To learn more about titration with the given link

brainly.com/question/2728613

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