__Zn + __HCl --> __ZnCl2 + __H2 If 26.22 Liters of Hydrogen gas are produced, how many moles of Hydrochloric acid were reacted?

1. The balanced equation:

Zn + 2HCl —> ZnCl₂ + H₂

The coefficients are: 1, 2, 1, 1

2. 2.34 moles of HCl

Explanation:

1. The balanced equation

__Zn + __HCl —> __ZnCl₂ + __H₂

There are 2 atoms of H on the right side and 1 atom on the left side

It can be balance by writing 2 before HCl as shown below:

Zn + 2HCl —> ZnCl₂ + H₂

Now the equation is balanced.

The coefficients are: 1, 2, 1, 1

2. Determination of the number of mole of HCl that reacted.

NOTE: 1 mole of H₂ = 22.4 L at STP.

Zn + 2HCl —> ZnCl₂ + H₂

From the balanced equation above,

2 moles of HCl reacted to produce 1 mole (i.e 22.4 L) of H₂.

Finally, we shall determine the number of mole of HCl needed to produce 26.22 L of H₂. This can be obtained as follow:

From the balanced equation above,

2 moles of HCl reacted to produce 22.4 L of H₂.

Therefore, Xmol of HCl will react to produce 26.22 L of H₂ i.e

Xmol of HCl = (2 × 26.22)/22.4

Xmol of HCl = 2.34 moles

Thus, 2.34 moles of HCl is needed for the reaction.

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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%