Calculate the mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid according to the balanced equation: 2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

The mass of hydrogen formed when 27 g of aluminum, Al reacts with excess hydrochloric acid, HCl is 3 g

We'll begin by calculating the mass of Al that reacted and the mass of H₂ formed from the balanced equation. This can be obtained as follow:

2Al + 6HCl → 2AlCl₃ + 3H₂

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 2 × 27 = 54 g

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 3 × 2 = 6 g

From the balanced equation above,

54 g of Al reacted to produce 6 g of H₂.

Finally, we shall determine the mass of H₂ produced by the reaction of 27 g of Al. This can be obtained as follow:

From the balanced equation above,

54 g of Al reacted to produce 6 g of H₂.

Therefore,

27 g of Al will react to produce = = 3 g of H₂.

Thus, the mass of H₂ produced from the reaction is 3 g

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6.069 grams is the  mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid.

Explanation:

Balanced equation for the reaction:

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

data given:

mass of aluminum = 27 grams

atomic mass of one mole of aluminum = 26.89 grams/mole

formula to calculate number of moles:

number of moles =

number of moles =

                              = 1.004 moles of aluminum will react

from the balanced equation:

2 moles of Al reacted to form 3 moles of H2

1.004 moles of Al will produce x moles of H2

=

x = 3.012 moles of H2 will be formed.

mass will be calculated as number of moles multiplied by atomic weight

mass of 3.012 moles of hydrogen ?(atomic weight of one mole H2 = 2.015 grams)

= 3.012 x 2.015

= 6.069 grams of H2 will be formed.

what are the óptions. pléase list them

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.