Suppose the following:
HO-C(H2)-C(H2)-OH + 2 HI --> I-C(H2)-C(H2)-I + water
How many mL of the first compound* are needed to react with 9.3 grams of HI? (The density of the first compound is A = 1.11 g/mL)
*by first compound, i mean HO-C(H2)-C(H2)-OH
To determine the volume in milliliters (mL) of the first compound (HO-C(H2)-C(H2)-OH) needed to react with 9.3 grams of HI, we can follow these steps:
Step 1: Convert the mass of HI to moles.
Using the molar mass of HI (127.91 g/mol), we can calculate the number of moles by dividing the mass by the molar mass.
Moles of HI = 9.3 g / 127.91 g/mol
Step 2: Use the stoichiometry of the balanced equation to find the mole ratio.
From the balanced equation, we can see that 1 mole of HO-C(H2)-C(H2)-OH reacts with 2 moles of HI.
Step 3: Convert moles of HI to moles of HO-C(H2)-C(H2)-OH.
Multiply the moles of HI obtained in step 1 by the mole ratio from step 2.
Moles of HO-C(H2)-C(H2)-OH = Moles of HI x (1 mole HO-C(H2)-C(H2)-OH / 2 moles HI)
Step 4: Convert moles of HO-C(H2)-C(H2)-OH to grams of HO-C(H2)-C(H2)-OH.
Multiply the moles of HO-C(H2)-C(H2)-OH obtained in step 3 by the molar mass of HO-C(H2)-C(H2)-OH.
Step 5: Convert grams of HO-C(H2)-C(H2)-OH to milliliters using density.
Divide the grams of HO-C(H2)-C(H2)-OH obtained in step 4 by the density of the compound (A = 1.11 g/mL).
Now, let's calculate it:
Step 1: Moles of HI = 9.3 g / 127.91 g/mol = 0.0726 mol
Step 2: Moles of HO-C(H2)-C(H2)-OH = Moles of HI x (1 mole HO-C(H2)-C(H2)-OH / 2 moles HI)
Moles of HO-C(H2)-C(H2)-OH = 0.0726 mol x (1 mol / 2 mol) = 0.0363 mol
Step 4: Grams of HO-C(H2)-C(H2)-OH = Moles of HO-C(H2)-C(H2)-OH x Molar mass of HO-C(H2)-C(H2)-OH.
To calculate the molar mass of HO-C(H2)-C(H2)-OH, we need to know the molecular formula and atomic masses of each element. Please provide that information so we can continue the calculation.
To find out how many mL of the first compound (HO-C(H2)-C(H2)-OH) are needed to react with 9.3 grams of HI, we first need to convert the mass of HI to moles. Then, we can use the balanced equation to determine the molar ratio between HI and HO-C(H2)-C(H2)-OH. Finally, we can convert the moles of HO-C(H2)-C(H2)-OH to mL using the density of the compound.
Let's start by calculating the moles of HI:
Step 1: Determine the molar mass of HI
The molar mass of hydrogen (H) is 1.01 g/mol, and the molar mass of iodine (I) is 126.90 g/mol. Therefore, the molar mass of HI is:
1.01 g/mol (H) + 126.90 g/mol (I) = 127.91 g/mol (HI)
Step 2: Convert grams of HI to moles
To convert grams of HI to moles, we use the formula:
moles = mass / molar mass
moles of HI = 9.3 g / 127.91 g/mol ≈ 0.0726 mol
Now, let's determine the molar ratio between HI and HO-C(H2)-C(H2)-OH using the balanced equation:
HO-C(H2)-C(H2)-OH + 2 HI → I-C(H2)-C(H2)-I + water
Based on the balanced equation, we can see that 1 mole of HO-C(H2)-C(H2)-OH reacts with 2 moles of HI.
Step 3: Calculate moles of HO-C(H2)-C(H2)-OH
moles of HO-C(H2)-C(H2)-OH = 0.0726 mol HI × (1 mol HO-C(H2)-C(H2)-OH / 2 mol HI)
= 0.0363 mol HO-C(H2)-C(H2)-OH
Finally, let's convert the moles of HO-C(H2)-C(H2)-OH to mL using the density of the compound:
Step 4: Use density to convert moles to mL
The density of HO-C(H2)-C(H2)-OH is given as A = 1.11 g/mL.
Step 5: Convert moles to grams
grams of HO-C(H2)-C(H2)-OH = moles of HO-C(H2)-C(H2)-OH × molar mass of HO-C(H2)-C(H2)-OH
molar mass of HO-C(H2)-C(H2)-OH = (1 × molar mass of H) + (4 × molar mass of C) + (8 × molar mass of H) + (1 × molar mass of O)
= (1 × 1.01 g/mol) + (4 × 12.01 g/mol) + (8 × 1.01 g/mol) + (1 × 16.00 g/mol)
= 44.11 g/mol
grams of HO-C(H2)-C(H2)-OH = 0.0363 mol HO-C(H2)-C(H2)-OH × 44.11 g/mol ≈ 1.60 g
Step 6: Convert grams to mL
mL of HO-C(H2)-C(H2)-OH = grams of HO-C(H2)-C(H2)-OH / density of HO-C(H2)-C(H2)-OH
mL of HO-C(H2)-C(H2)-OH = 1.60 g / 1.11 g/mL ≈ 1.44 mL (rounded to two decimal places)
Therefore, approximately 1.44 mL of the first compound (HO-C(H2)-C(H2)-OH) are needed to react with 9.3 grams of HI.