Consider the following chemical equation.

H2(g) + I2(g) → HI(g)
Assuming all gases are at the same temperature and pressure, how many milliliters of hydrogen iodide are produced from 825 mL of H2?

To determine the volume of hydrogen iodide (HI) produced from 825 mL of hydrogen gas (H2), we need to use the ideal gas law and stoichiometry.

First, let's write a balanced chemical equation for the reaction:
H2(g) + I2(g) → 2HI(g)

From the balanced equation, we can see that one mole of hydrogen gas (H2) reacts to produce two moles of hydrogen iodide (HI).

Step 1: Convert mL of H2 to moles of H2
To do this, we need to know the molar volume of an ideal gas at standard temperature and pressure (STP), which is 22.4 L/mol.

1 L = 1000 mL (Conversion factor)
22.4 L = 22.4 × 1000 mL = 22,400 mL (Conversion factor)

So, 22,400 mL of an ideal gas at STP is equivalent to 1 mole.

To calculate the number of moles of H2 in 825 mL, we use the following equation:

moles of H2 = (volume of H2 in mL) / (molar volume of H2 at STP)

moles of H2 = 825 mL / 22,400 mL/mol

Step 2: Use stoichiometry to find the moles of HI produced
From the balanced equation, we know that 1 mole of H2 produces 2 moles of HI.

moles of HI = (moles of H2) × (2 moles of HI / 1 mole of H2)

Step 3: Convert moles of HI to volume in mL
To convert moles of HI to volume in mL, we use the molar volume of an ideal gas at STP (22.4 L/mol) again.

volume of HI = (moles of HI) × (molar volume of HI at STP)

volume of HI = (moles of HI) × (22.4 L/mol) × (1000 mL/L)

Now you can plug in the values and calculate the answer.