Assuming all gases are at the same temperature and pressure, how many milliliters of hydrogen iodide are produced from 165 mL of H2?
H2(g)+I2(g)→2HI(g)
Doesn;t it depend upon how much I2 you have initially.
o=Ok. How to solve it? Thanks.
To solve this problem, we can use the concept of stoichiometry, which allows us to relate the amounts of reactants and products in a chemical reaction.
First, let's write and balance the chemical equation:
H2(g) + I2(g) → 2HI(g)
The balanced chemical equation tells us that one molecule of H2 reacts with one molecule of I2 to produce two molecules of HI.
To find out how many milliliters of hydrogen iodide (HI) are produced from 165 mL of H2, we need to calculate the volume of HI using stoichiometry.
Step 1: Convert the given volume of H2 to the number of moles.
To do this, we need to know the molar volume of H2, which is 22.4 L/mol at standard temperature and pressure (STP). Since we're given the volume in milliliters, we need to convert it to liters:
165 mL ÷ 1000 mL/L = 0.165 L
Now we can calculate the number of moles of H2:
0.165 L × (1 mol H2 / 22.4 L) = 0.00735 mol H2
Step 2: Using the stoichiometry of the balanced equation, determine the number of moles of HI produced.
According to the stoichiometry, for every 1 mole of H2, 2 moles of HI are produced.
So, the number of moles of HI can be calculated as follows:
0.00735 mol H2 × (2 mol HI / 1 mol H2) = 0.0147 mol HI
Step 3: Convert the number of moles of HI to volume in milliliters.
To do this, we need to know the molar volume of HI, also at STP. We can use the same value as for H2, which is 22.4 L/mol.
0.0147 mol HI × (22.4 L / 1 mol HI) × (1000 mL / 1 L) = 329 mL HI (rounded to the nearest whole number)
Therefore, approximately 329 mL of hydrogen iodide (HI) are produced from 165 mL of H2, assuming all gases are at the same temperature and pressure.