Answer:
32.7 L of H₂
Explanation:
We'll begin by calculating the number of mole in 95 g of Zn. This can be obtained as follow:
Mass of Zn = 95 g
Molar mass of Zn = 65 g/mol
Mole of Zn =?
Mole = mass /Molar mass
Mole of Zn = 95/65
Mole of Zn = 1.46 mole
Next, we shall determine the number of mole of H₂ produced by the reaction of 95 g (i.e 1.46 mole) of Zn. This can be obtained as illustrated below:
Zn + H₂SO₄ —> ZnSO₄ + H₂
From the balanced equation above,
1 mole of Zn reacted to produce 1 mole of H₂.
Therefore, 1.46 mole of Zn will also react to produce 1.46 mole of H₂.
Finally, we shall determine the volume of H₂ obtained at STP. This can be obtained as follow:
1 mole of H₂ = 22.4 L at STP.
Therefore,
1.46 mole of H₂ = 1.46 × 22.4
1.46 mole of H₂ = 32.7 L at STP
Thus, 32.7 L of H₂ were obtained from the reaction.
