What volume of H2 could be produced at a temperature of 35 oC and a pressure of 769 torr by the reaction of 2.82 g of aluminum with hydrochloric acid?

3.9 L of H2.

Explanation:

We'll begin by calculating the number of mole in 2.82 g of aluminium, Al.

This is illustrated below:

Mass of Al = 2.82 g

Molar mass of Al = 27 g/mol

Mole of Al =?

Mole = mass /Molar mass

Mole of Al = 2.82 / 27

Mole of Al = 0.104 mole

Next, we shall write the balanced equation for the reaction. This is given below:

2Al + 6HCI —> 2AlCl3 + 3H2

From the balanced equation above,

2 moles of Al reacted with 6 moles of HCl to produce 3 moles of H2.

Next, we shall determine the number of mole H2 produced by reacting 0.104 mole of Al.

This is illustrated below:

From the balanced equation above,

2 moles of Al reacted to produce 3 moles of H2.

Therefore, 0.104 mole of Al will react to produce = (0.104 x 3)/2 = 0.156 mole of H2.

Therefore, 0.156 mole of H2 was produced from the reaction.

Finally, we shall determine the volume of H2 produced by using the ideal gas equation as shown below:

Number of mole (n) of H2 = 0.156 mole

Pressure (P) = 769 torr = 769/760 = 1.01 atm

Temperature (T) = 35°C = 35°C + 273 = 308 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) of H2 =?

PV = nRT

1.01 x V = 0.156 x 0.0821 x 308

Divide both side by 1.01

V = (0.156 x 0.0821 x 308) /1.01

V = 3.9 L

Therefore, 3.9 L of H2 were produced from the reaction.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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