A scientist studying the reaction between decaborane (B10H18) and oxygen mixed 1.00 x 10^2 grams fo oxygen. Compute the masses of all substances present after the reaction went to completion, generating B2O3 and water as the only products.
Silicon carbide, SiC, is prepared by hearing silicon dioxide in the presence of graphite. Carbon dioxide is the by-product of the reaction. How many grams of silicon carbide can be formed from the reaction of 50.0 grams of graphite with 50.0 grams of silicone dioxide?
Here is what I have so far
SiO2 + 3C > SiC + CO2
Well, this sounds like quite the explosive experiment! Let's break it down shall we?
First, let's find the molar mass of decaborane (B10H18). By adding up the atomic masses of each element, we get a molar mass of approximately 122.7 g/mol.
Now, let's calculate how many moles of oxygen we have. We have 1.00 x 10^2 grams of oxygen, so we divide that by the molar mass of oxygen (16 g/mol) and get approximately 6.25 moles of oxygen.
The reaction between decaborane and oxygen will produce B2O3 and water. Since the balanced equation is not provided, let's assume that for every two moles of decaborane, we produce one mole of B2O3 and nine moles of water.
Therefore, if we have 6.25 moles of oxygen, we must have half that amount, or 3.125 moles, of decaborane to react completely.
So, the mass of decaborane present after the reaction is 3.125 moles multiplied by the molar mass of decaborane (122.7 g/mol), which is approximately 384 grams.
For the products, we get one mole of B2O3 and nine moles of water for every two moles of decaborane. Therefore, we have approximately 1.56 moles of B2O3 and 14.06 moles of water.
The mass of B2O3 is 1.56 moles multiplied by the molar mass of B2O3 (69.6 g/mol), which is approximately 108.58 grams.
The mass of water is 14.06 moles multiplied by the molar mass of water (18 g/mol), which is approximately 253.08 grams.
So, after the reaction went to completion, we have approximately:
- 384 grams of decaborane
- 108.58 grams of B2O3
- 253.08 grams of water
Hope that helps! Just be careful not to create any fire hazards while conducting these chemical reactions, or you might end up with a chemistry-themed fireworks display!
To solve this problem, we need to balance the chemical equation for the reaction between decaborane (B10H18) and oxygen (O2) to form B2O3 and water (H2O).
The balanced equation is:
2B10H18 + 21O2 -> B2O3 + 18H2O
From this equation, we can see that 2 moles of decaborane react with 21 moles of oxygen to produce 1 mole of B2O3 and 18 moles of water.
Before we can calculate the masses of the substances present after the reaction, we need to determine the moles of oxygen used in the reaction.
Given:
Mass of oxygen = 1.00 x 10^2 grams
Molar mass of oxygen (O2) = 32.00 g/mol
To calculate the moles of oxygen, we can use the formula:
moles = mass / molar mass
moles of oxygen = 1.00 x 10^2 g / 32.00 g/mol
moles of oxygen ≈ 3.13 mol
Now, let's calculate the moles of other substances using the mole ratios from the balanced equation.
From the balanced equation, we have the following mole ratios:
2 moles of decaborane react with 21 moles of oxygen to produce 1 mole of B2O3
2 moles of decaborane react with 21 moles of oxygen to produce 18 moles of water
Since the reaction goes to completion, all the decaborane will react.
moles of decaborane = 2 * moles of oxygen
moles of decaborane = 2 * 3.13 mol
moles of decaborane = 6.26 mol
Therefore, after the reaction is complete, we have:
- Moles of B2O3 produced = 1 * moles of decaborane
- Moles of water produced = 18 * moles of decaborane
Now, we can calculate the masses of the substances present:
Given:
Molar mass of B2O3 = 69.62 g/mol
Molar mass of water (H2O) = 18.02 g/mol
Mass of B2O3 = Moles of B2O3 * Molar mass of B2O3
Mass of B2O3 = 1 * 6.26 mol * 69.62 g/mol
Mass of B2O3 ≈ 435.94 g
Mass of water = Moles of water * Molar mass of water
Mass of water = 18 * 6.26 mol * 18.02 g/mol
Mass of water ≈ 2001.84 g
Therefore, after the reaction went to completion, the masses of the substances present are approximately:
- Mass of B2O3 = 435.94 grams
- Mass of water = 2001.84 grams
To compute the masses of all substances present after the reaction, we'll need to balance the chemical equation for the reaction first. The reaction between decaborane (B10H18) and oxygen (O2) can be represented by the following equation:
2 B10H18 + 21 O2 → 10 B2O3 + 18 H2O
From the balanced equation, we can see that 2 moles of B10H18 react with 21 moles of O2 to produce 10 moles of B2O3 and 18 moles of H2O.
Now, let's calculate the molar masses of the substances involved in the reaction:
Molar mass of B10H18:
(10 × 10 atomic mass units of boron) + (18 × 1 atomic mass unit of hydrogen) = 180 g/mol
Molar mass of O2:
(2 × 16 atomic mass units of oxygen) = 32 g/mol
Molar mass of B2O3:
(2 × 10 atomic mass units of boron) + (3 × 16 atomic mass units of oxygen) = 70 g/mol
Molar mass of H2O:
(2 × 1 atomic mass unit of hydrogen) + (16 atomic mass unit of oxygen) = 18 g/mol
Now, let's calculate the number of moles of oxygen present in the reaction:
Given mass of oxygen = 1.00 × 10^2 grams
Molar mass of O2 = 32 g/mol
Number of moles of O2 = (Given mass of oxygen) / (Molar mass of O2) = (1.00 × 10^2 g) / (32 g/mol) = 3.125 mol
According to the stoichiometry of the equation, 2 moles of B10H18 react with 21 moles of O2. Therefore, if 21 moles of O2 react, we will need to scale the reaction to determine the corresponding amounts of B10H18, B2O3, and H2O produced.
Number of moles of B10H18 = (2 moles B10H18 / 21 moles O2) × (3.125 moles O2) = 0.2976 mol
Now, we can calculate the masses of the substances present after the reaction:
Mass of B10H18 = (Molar mass of B10H18) × (Number of moles of B10H18) = (180 g/mol) × (0.2976 mol) ≈ 53.568 g
Mass of B2O3 = (Molar mass of B2O3) × (Number of moles of B2O3) = (70 g/mol) × (10 moles / 21 moles of O2) × (3.125 mol) ≈ 93.75 g
Mass of H2O = (Molar mass of H2O) × (Number of moles of H2O) = (18 g/mol) × (18 moles / 21 moles of O2) × (3.125 mol) ≈ 47.771 g
Therefore, after the reaction went to completion, the mass of decaborane (B10H18) would be approximately 53.568 grams, the mass of B2O3 would be approximately 93.75 grams, and the mass of H2O would be approximately 47.771 grams.