Type Calculations. Given the balanced equation: 2 Al + 3 H2SO4---> Al2(SO4)3 + 3 H2 Molar mass (g/mol): Al=26.98; H2SO4=98.08; Al2(SO4)3= 342.15; H2=2.02 A) How many mole of H2 gas can be produced when 33.8 g of Al are consumed? B) When 1.60 mol of H2SO4 are used in a reaction, how many grams of Al2(SO4)3 can be produced? C) For part B, if actual yield for Al2(SO4)3 is 100.0 g, what is percent yield?

A.  1.88 mol H₂

B.  182 g Al₂(SO₄)₃

C.  54.8%

Explanation:

2 Al  +  3 H₂SO₄  ⇒  Al₂(SO₄)₃  +  3 H₂

A.  Convert grams of Al to moles.  The molar mass is 26.98 g/mol.

(33.8 g)/(26.98 g/mol) = 1.253 mol Al

Use stoichiometry to convert moles of Al to moles of H₂.  Looking at the equation, you can see that for every 2 mol of Al consumed, 3 moles of H₂ is produced.  Use this relationship.

(1.253 mol Al) × (3 mol H₂)/(2 mol Al) = 1.879 mol H₂

You will produce 1.88 mol of H₂ gas.

B.  Again, use stoichiometry.  For every 3 moles of H₂SO₄ consumed, 1 mole of Al₂(SO₄)₃ is produced.

(1.60 mol H₂SO₄) × (1 mol Al₂(SO₄)₃/3 mol H₂SO₄) = 0.533 mol Al₂(SO₄)₃

Convert moles of Al₂(SO₄)₃ to grams.  The molar mass is 342.15 g/mol.

(0.533 mol) × (342.15 g/mol) = 182.48 g Al₂(SO₄)₃

You will produce 182 g of Al₂(SO₄)₃.

C.  Calculate percent yield by dividing the actual yield by the theoretical yield.  Multiply by 100%.

(100.0/182.48) × 100% = 54.8%

The percent yield is 54.8%.

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.