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What volume of hydrogen will be produced at STP by the reaction of 67.3 g of tin with excess water according to the following reaction?

Respuesta :

itzmeshlok

Answer:

Sn + 2H2O ==> Sn(OH)2 + 2H2

67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters

Explanation:

Sn + 2H2O ==> Sn(OH)2 + 2H2

67.3 g Sn x 1 mol/119 g x 2 mol H2/mol Sn x 22.4 L/mole = answer in liters

Eduard22sly

The volume of hydrogen that will be produced from the reaction of 67.3 g of tin with excess water is 12.68 L

How to determine the mole of tin

  • Mass of Sn = 67.3 g
  • Molar mass of Sn = 119 g/mol
  • Mole of Sn =?

Mole = mass / molar mass

Mole of Sn = 67.3 / 119

Mole of Sn = 0.566 mole

How to determine the volume of H₂ produced

Sn + 2H₂O —> Sn(OH)₂ + H₂

From the balanced equation above,

1 mole of Sn reacted to produce 1 mole of H₂.

Therefore,

0.566 mole of Sn will also react to produce 0.566 mole of H₂

Recall

1 mole of H₂ = 22.4 L at STP

Therefore,

0.566 mole of H₂ = 0.566 × 22.4

0.566 mole of H₂ = 12.68 L

Thus, 12.68 L of H₂ were obtained from the reaction.

Learn more about stoichiometry:

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