How many grams of H2(g)can theoretically be produced from a 5-g sample of magnesium metal and excess HC1(aq)?

Here is an example stoichiometry problem I've posted. Just follow the steps.

oops. Forgot the link. Here is an example stoichiometry problem I've posted. Just follow the steps. http://www.jiskha.com/display.cgi?id=1289515206

To determine the amount of H2(g) produced, we need to use stoichiometry and the balanced chemical equation.

The balanced chemical equation for the reaction between magnesium (Mg) and hydrochloric acid (HCl) is:
Mg + 2HCl → MgCl2 + H2

From the equation, we can see that 1 mole of magnesium (24.31 g) reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas (2.02 g).

Now, let's calculate the moles of magnesium present in the 5 g sample using the molar mass of magnesium:
Moles of Mg = Mass of Mg / Molar mass of Mg
Moles of Mg = 5 g / 24.31 g/mol
Moles of Mg = 0.206 mol

Since the reaction ratio is 1:1 between magnesium and hydrogen gas, the moles of hydrogen gas produced will be equal to the moles of magnesium.

Therefore, the amount of hydrogen gas produced from the 5 g sample of magnesium metal is 0.206 mol.

To convert moles to grams, we use the molar mass of hydrogen gas:
Mass of H2 = Moles of H2 x Molar mass of H2
Mass of H2 = 0.206 mol x 2.02 g/mol
Mass of H2 = 0.416 g

Theoretically, approximately 0.416 grams of H2(g) can be produced from a 5 g sample of magnesium metal and excess HCl(aq).

To calculate the number of grams of H2(g) that can be produced from a given sample of magnesium metal and excess HCl(aq), we need to use stoichiometry.

Step 1: Write the balanced chemical equation for the reaction:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Step 2: Determine the molar mass of magnesium (Mg) and hydrogen (H2).
Molar mass of Mg = 24.31 g/mol
Molar mass of H2 = 2.02 g/mol

Step 3: Calculate the number of moles of magnesium in the 5 g sample.
Number of moles of Mg = Mass of Mg / Molar mass of Mg
Number of moles of Mg = 5 g / 24.31 g/mol

Step 4: Use the stoichiometry of the balanced equation to calculate the number of moles of H2 produced.
From the balanced equation, we can see that 1 mole of Mg reacts to produce 1 mole of H2.
Number of moles of H2 = Number of moles of Mg * (1 mole of H2 / 1 mole of Mg)

Step 5: Calculate the mass of H2 produced.
Mass of H2 = Number of moles of H2 * Molar mass of H2

So, the total mass of H2 that can theoretically be produced from the 5 g sample of magnesium metal is given by:
Mass of H2 = Number of moles of H2 * Molar mass of H2

Let's calculate it step by step:
Step 1: Balanced chemical equation:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Step 2: Molar mass of Mg = 24.31 g/mol
Molar mass of H2 = 2.02 g/mol

Step 3: Number of moles of magnesium (Mg):
Number of moles of Mg = Mass of Mg / Molar mass of Mg
Number of moles of Mg = 5 g / 24.31 g/mol

Step 4: Number of moles of H2:
Number of moles of H2 = Number of moles of Mg * (1 mole of H2 / 1 mole of Mg)

Step 5: Mass of H2:
Mass of H2 = Number of moles of H2 * Molar mass of H2

Now, let's calculate it:

Number of moles of Mg = 5 g / 24.31 g/mol
Number of moles of Mg ≈ 0.206 moles of Mg

Number of moles of H2 = 0.206 moles of Mg * (1 mole of H2 / 1 mole of Mg)
Number of moles of H2 ≈ 0.206 moles of H2

Mass of H2 = 0.206 moles of H2 * 2.02 g/mol
Mass of H2 ≈ 0.417 g of H2

Therefore, theoretically, approximately 0.417 grams of H2 gas can be produced from a 5-gram sample of magnesium metal and excess HCl(aq).