A scientist studying the reaction between decaborane (B10H18) and oxygen mixed 1.00 x 10^2 grams of decaborane with 2.75 x 10^2 grams of oxygen. Compute the masses of all substance present after the reaction went ot completion, generating B2O2 and water as the only products.
To determine the masses of all substances present after the reaction, we need to calculate the stoichiometric ratios between the reactants and products.
First, let's write the balanced chemical equation for the reaction between decaborane (B10H18) and oxygen (O2):
2B10H18 + 21O2 -> 15B2O2 + 18H2O
From the balanced equation, we can see that 2 moles of decaborane react with 21 moles of oxygen to produce 15 moles of B2O2 and 18 moles of water.
Next, we need to convert the given masses of decaborane (B10H18) and oxygen (O2) to moles.
The molar mass of decaborane (B10H18) can be calculated as follows:
10 atoms of boron (B) × atomic mass of boron (10.81 g/mol) = 108.1 g/mol
18 atoms of hydrogen (H) × atomic mass of hydrogen (1.01 g/mol) = 18.18 g/mol
Total molar mass of decaborane = 108.1 g/mol + 18.18 g/mol = 126.28 g/mol
To convert the mass of decaborane to moles, we use the formula:
moles = mass / molar mass
Number of moles of decaborane = 1.00 x 10^2 g / 126.28 g/mol
Similarly, the molar mass of oxygen (O2) is 32.00 g/mol.
Number of moles of oxygen = 2.75 × 10^2 g / 32.00 g/mol
Now, we need to determine the limiting reactant, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed. The reactant that has the smaller number of moles will be the limiting reactant.
To find the limiting reactant, we compare the number of moles of decaborane and oxygen:
moles of decaborane = 1.00 x 10^2 g / 126.28 g/mol = x1
moles of oxygen = 2.75 × 10^2 g / 32.00 g/mol = x2
Since there are more moles of oxygen (x2) than decaborane (x1), decaborane is the limiting reactant.
Now, we can calculate the moles of products formed based on the stoichiometric ratios from the balanced equation.
Moles of B2O2 = 15 × x1
Moles of H2O = 18 × x1
Finally, we need to convert the moles of B2O2 and H2O back to mass:
Mass of B2O2 = Moles of B2O2 × molar mass of B2O2
Mass of H2O = Moles of H2O × molar mass of H2O
By following these calculations, we can determine the masses of all substances present after the reaction has gone to completion.
To determine the masses of all substances present after the reaction, we'll follow these steps:
Step 1: Write the Balanced Chemical Equation
Step 2: Determine the Limiting Reactant
Step 3: Calculate the Moles of Products Formed
Step 4: Convert Moles to Grams
Let's go through each step one by one.
Step 1: Write the Balanced Chemical Equation
Decaborane (B10H18) reacts with oxygen (O2) to form B2O2 (diboron dioxide) and water (H2O).
The balanced chemical equation for the reaction is:
2B10H18 + 11O2 → 5B2O2 + 9H2O
Step 2: Determine the Limiting Reactant
To find the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
The molar mass of decaborane (B10H18) is 122.53 g/mol.
The molar mass of oxygen (O2) is 32.00 g/mol.
Calculate the moles of each reactant:
Moles of decaborane = mass of decaborane / molar mass of decaborane
= 100.00 g / 122.53 g/mol
= 0.815 mol
Moles of oxygen = mass of oxygen / molar mass of oxygen
= 275.00 g / 32.00 g/mol
= 8.59 mol
According to the balanced equation, the stoichiometric ratio between decaborane and oxygen is 2:11. Therefore, the limiting reactant is decaborane because it is present in a smaller quantity.
Step 3: Calculate the Moles of Products Formed
From the balanced equation, we can see that 2 moles of decaborane will produce 5 moles of B2O2 and 9 moles of H2O.
Since decaborane is the limiting reactant, it will be completely consumed, and we can calculate the moles of products formed based on the stoichiometry.
Moles of B2O2 produced = 5 moles B2O2 / 2 moles decaborane * moles of decaborane
= 5/2 * 0.815 mol
= 2.038 mol
Moles of H2O produced = 9 moles H2O / 2 moles decaborane * moles of decaborane
= 9/2 * 0.815 mol
= 3.659 mol
Step 4: Convert Moles to Grams
Finally, we'll convert the moles of each substance back to grams.
Mass of B2O2 = Moles of B2O2 * molar mass of B2O2
= 2.038 mol * (11.81 g/mol + 15.999 g/mol)
= 71.35 g
Mass of H2O = Moles of H2O * molar mass of H2O
= 3.659 mol * (2.016 g/mol + 15.999 g/mol)
= 69.15 g
Therefore, the masses of all substances present after the reaction went to completion are:
- 71.35 grams of B2O2 (diboron dioxide)
- 69.15 grams of H2O (water)