If a student working on the upcoming experiment fully ionizes 2.80 g of Zn(s) in an excess of aqueous HCl, how many moles of H2 (g) will be produced?

Here is a worked example of a stoichiometry problem. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine the number of moles of H2 gas produced when 2.80 g of Zn(s) is fully ionized in an excess of aqueous HCl, we need to follow a step-by-step approach.

Step 1: Write and balance the chemical equation
The chemical equation for the reaction between zinc (Zn) and hydrochloric acid (HCl) can be written as follows:

Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)

Step 2: Calculate the molar mass of Zn
The molar mass of Zn can be found by looking up the atomic mass of zinc from the periodic table. The atomic mass of zinc (Zn) is approximately 65.38 g/mol.

Step 3: Convert the mass of Zn to moles
To convert the mass of Zn to moles, you can use the formula:

moles = mass / molar mass

In this case, the mass of Zn is given as 2.80 g. Plugging in the values, we have:

moles of Zn = 2.80 g / 65.38 g/mol

Step 4: Determine the stoichiometry of the reaction
From the balanced chemical equation, we can see that 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of H2 gas. This means the mole ratio of Zn to H2 is 1:1.

Step 5: Calculate the moles of H2 produced
Since the mole ratio of Zn to H2 is 1:1, the number of moles of H2 produced is the same as the moles of Zn. Therefore, the moles of H2 can be determined as:

moles of H2 = moles of Zn

Now, you can substitute the value of moles of Zn obtained in Step 3 to calculate the moles of H2 gas produced when 2.80 g of Zn(s) is fully ionized in an excess of aqueous HCl.

It is important to note that the given mass of Zn (2.80 g) is assumed to be fully ionized in this calculation.