Silicon Carbide, SiC, is prepared by heating 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 fo graphite with 50.0 grams of silicon dioxide?
This looks like a limiting reagent (LR) problem.
SiO2 + 2C ==> SiC + CO2
mols SiO2 = grams/molar mass = approx 50/60.1 = approx 0.0832
mols C = 50/12 = approx 0.0416
mols SiC formed using just SiO2 = 1:1; therefore, 0.0832.
mols SiC formed using just C = 0.0416 x 1/2 = 0.0208 so C is the limiting reagent and the amount of SiC formed is the SMALLER of the two possible answers.
grams SiC = mols SiC x molar mass SiC = ?
To determine the amount of silicon carbide that can be formed from the reaction, we need to use the concept of stoichiometry. Stoichiometry is the calculation of quantitative relationships between reactants and products in a chemical reaction.
1. Write the balanced equation for the reaction:
SiO2 + 3C -> SiC + 2CO2
From the equation, we can see that 1 mole of silicon dioxide (SiO2) reacts with 3 moles of carbon (C) to produce 1 mole of silicon carbide (SiC) and 2 moles of carbon dioxide (CO2).
2. Calculate the number of moles of reactants:
Number of moles of graphite (C) = Mass of graphite / Molar mass of graphite
Number of moles of graphite (C) = 50.0 g / 12.01 g/mol = 4.162 moles
Number of moles of silicon dioxide (SiO2) = Mass of silicon dioxide / Molar mass of silicon dioxide
Number of moles of silicon dioxide (SiO2) = 50.0 g / 60.08 g/mol = 0.833 moles
3. Determine the limiting reactant:
To find the limiting reactant, compare the mole ratios of the reactants in the balanced equation. The reactant that produces the least amount of product is the limiting reactant.
From the balanced equation, we can see that 3 moles of carbon react with 1 mole of silicon dioxide to produce 1 mole of silicon carbide. Therefore, the ratio of moles of carbon to moles of silicon dioxide is 3:1.
Using this ratio, we can calculate the moles of silicon dioxide that can react with 4.162 moles of carbon:
Moles of silicon dioxide = Moles of carbon x (1 mole of silicon dioxide / 3 moles of carbon)
Moles of silicon dioxide = 4.162 moles x (1 mole of silicon dioxide / 3 moles of carbon) = 1.387 moles
Since we have 0.833 moles of silicon dioxide available and 1.387 moles of silicon dioxide are needed, silicon dioxide is the limiting reactant.
4. Calculate the amount of silicon carbide formed:
From the balanced equation, 1 mole of silicon dioxide reacts to form 1 mole of silicon carbide. Therefore, the moles of silicon carbide formed are the same as the moles of silicon dioxide used.
Moles of silicon carbide formed = Moles of silicon dioxide used = 0.833 moles
5. Finally, calculate the mass of silicon carbide formed:
Mass of silicon carbide = Moles of silicon carbide formed x Molar mass of silicon carbide
Mass of silicon carbide = 0.833 moles x 40.10 g/mol = 33.39 grams
Therefore, approximately 33.39 grams of silicon carbide can be formed from the reaction of 50.0 grams of graphite with 50.0 grams of silicon dioxide.
To calculate the grams of silicon carbide formed, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed, and it determines the maximum amount of product that can be formed.
Step 1: Convert the mass of each reactant to moles using their respective molar masses.
- Molar mass of graphite (C) = 12.01 g/mol
- Molar mass of silicon dioxide (SiO2) = 60.08 g/mol
Moles of graphite = mass of graphite / molar mass of carbon = 50.0 g / 12.01 g/mol = 4.164 mol
Moles of silicon dioxide = mass of SiO2 / molar mass of SiO2 = 50.0 g / 60.08 g/mol = 0.832 mol
Step 2: Determine the stoichiometric ratio of graphite and silicon carbide from the balanced chemical equation.
The balanced equation for the reaction is:
SiO2 + 3C -> SiC + 2CO2
From the equation, we can see that 1 mole of silicon dioxide reacts with 3 moles of carbon to form 1 mole of silicon carbide. Therefore, the stoichiometric ratio is:
1 mol SiO2 : 3 mol C : 1 mol SiC
Step 3: Determine the limiting reagent.
Compare the moles of the two reactants to their respective stoichiometric ratios. The reactant with the smaller ratio is the limiting reagent.
For graphite:
4.164 mol C / 3 mol SiC = 1.388 mol
For silicon dioxide:
0.832 mol SiO2 / 1 mol SiC = 0.832 mol
Since the moles of graphite are greater than the moles of silicon dioxide, the limiting reagent is silicon dioxide.
Step 4: Calculate the moles and grams of silicon carbide formed.
From the balanced equation, we know that 1 mole of silicon dioxide reacts to form 1 mole of silicon carbide.
Therefore, the number of moles of silicon carbide formed is equal to the moles of silicon dioxide.
Moles of SiC formed = 0.832 mol
Finally, convert the moles of silicon carbide formed to grams using its molar mass:
Molar mass of silicon carbide (SiC) = 40.10 g/mol
Mass of SiC formed = Moles of SiC formed x Molar mass of SiC
= 0.832 mol x 40.10 g/mol = 33.35 g
Therefore, 33.35 grams of silicon carbide can be formed from the reaction of 50.0 grams of graphite with 50.0 grams of silicon dioxide.