Silicon carbide, commonly known as carborundum, is a very hard and abrasive substance. The compound is prepared by heating silicon dioxide in the presence of graphitic carbon:

SiO2(s) + 2C(s) --> SiC(s) + CO2(g)

How many grams of silicon carbide can be formed from 20.0 g of graphite and 65.0 g of silicon dioxide?

Well, let's see if silicon carbide can measure up!

To find out how much silicon carbide can be formed, we need to determine the limiting reactant. This is the reactant that will be completely used up, while the other reactant will have some left over.

First, let's calculate the moles of each reactant using their molar masses.

The molar mass of graphite (C) is approximately 12.01 g/mol, so 20.0 g of graphite would be equal to 20.0 g / 12.01 g/mol = 1.67 mol.

The molar mass of silicon dioxide (SiO2) is approximately 60.08 g/mol, so 65.0 g of silicon dioxide would be equal to 65.0 g / 60.08 g/mol = 1.08 mol.

The balanced equation tells us that 2 moles of graphite react with 1 mole of silicon dioxide to form 1 mole of silicon carbide.

Therefore, if 1.67 mol of graphite reacts, we would need 0.835 mol of silicon dioxide to react completely. However, we only have 1.08 mol available, which means silicon dioxide is in excess.

So, silicon dioxide is the limiting reactant, and we can calculate the amount of silicon carbide formed based on the amount of silicon dioxide used.

From the balanced equation, we know that 1 mole of silicon dioxide reacts to form 1 mole of silicon carbide. Therefore, the amount of silicon carbide formed will be 1.08 mol.

Now, let's calculate the mass of silicon carbide formed using its molar mass.

The molar mass of silicon carbide (SiC) is approximately 40.10 g/mol. So, 1.08 mol of silicon carbide would be equal to 1.08 mol * 40.10 g/mol = 43.23 g.

Therefore, from 65.0 g of silicon dioxide and 20.0 g of graphite, we can form approximately 43.23 grams of silicon carbide.

So, the answer is: approximately 43.23 grams of silicon carbide can be formed. Time to carbide-load and get grinding!

To determine the amount of silicon carbide formed, we need to do a stoichiometric calculation based on the given balanced equation:

SiO2(s) + 2C(s) --> SiC(s) + CO2(g)

The molar mass of silicon dioxide (SiO2) is 60.08 g/mol, and the molar mass of graphite (C) is 12.01 g/mol.

Step 1: Calculate the moles of graphite (C) and silicon dioxide (SiO2) using their respective molar masses.

Moles of C = Mass of graphite / Molar mass of graphite
= 20.0 g / 12.01 g/mol
= 1.665 mol (rounded to three decimal places)

Moles of SiO2 = Mass of silicon dioxide / Molar mass of SiO2
= 65.0 g / 60.08 g/mol
= 1.082 mol (rounded to three decimal places)

Step 2: Based on the balanced equation, the stoichiometric ratio between graphite (C) and silicon carbide (SiC) is 2:1. This means that for every 2 moles of graphite, we get 1 mole of silicon carbide.

Step 3: Determine the limiting reagent by comparing the moles of graphite and silicon dioxide. The reactant that produces fewer moles of silicon carbide will be the limiting reagent.

Since 2 moles of graphite produce 1 mole of silicon carbide, the maximum moles of SiC that can be formed from graphite = (1.665 mol graphite / 2) = 0.8325 mol SiC.

Since 1 mole of silicon dioxide produces 1 mole of silicon carbide, the maximum moles of SiC that can be formed from SiO2 = 1.082 mol SiC.

The limiting reagent is graphite because it produces fewer moles of silicon carbide than silicon dioxide.

Step 4: Calculate the mass of silicon carbide formed using the moles of the limiting reagent and the molar mass of SiC, which is 40.10 g/mol.

Mass of SiC = Moles of SiC (limiting reagent) × Molar mass of SiC
= 0.8325 mol × 40.10 g/mol
= 33.34 g (rounded to two decimal places)

Therefore, 33.34 grams of silicon carbide can be formed from 20.0 grams of graphite and 65.0 grams of silicon dioxide.

To determine the amount of silicon carbide that can be formed, we need to use stoichiometry. Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction.

1. Start by writing down the balanced chemical equation for the reaction:
SiO2(s) + 2C(s) --> SiC(s) + CO2(g)

2. Convert the given masses of graphite (C) and silicon dioxide (SiO2) into moles. To do this, divide each mass by the molar mass of the substance. The molar mass of graphite (C) is approximately 12.01 g/mol, and the molar mass of silicon dioxide (SiO2) is approximately 60.08 g/mol.

Moles of graphite (C) = 20.0 g / 12.01 g/mol
Moles of silicon dioxide (SiO2) = 65.0 g / 60.08 g/mol

3. Determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, thus determining the maximum amount of product that can be formed. To find the limiting reactant, compare the mole ratios from the balanced equation to calculate the number of moles of silicon carbide that can be formed from each reactant.

From the balanced equation, we know that:
1 mole of SiO2 reacts to form 1 mole of SiC
2 moles of C react to form 1 mole of SiC

Moles of SiC from SiO2 = Moles of SiO2
Moles of SiC from C = (Moles of C) / 2

Compare the calculated moles of SiC from each reactant. The smaller value will indicate the limiting reactant.

4. Once you have identified the limiting reactant, calculate the maximum moles of SiC that can be formed using the stoichiometry of the balanced equation. The balanced equation tells us the mole ratio between reactants and products.

Moles of SiC = (Moles of limiting reactant) * (Mole ratio of SiC to limiting reactant)

5. Finally, convert the moles of SiC to grams using the molar mass of SiC, which is approximately 40.10 g/mol.

Grams of SiC = Moles of SiC * Molar mass of SiC

By following these steps, you will be able to calculate the grams of silicon carbide that can be formed from 20.0 g of graphite and 65.0 g of silicon dioxide.

A limiting reagent (LR) problem. You know that when you are given amounts for BOTH reactants.

mols C = 20/12 = approx 1.7 but you need a more accurate answer for this and all of the estimates that follow.

mols Si = g/atomic mass = 65/60 = about 1

mols SiC formed if you had 1.7 mols C and all of the SiO2 needed is 1.7 x (1 mol SiC/2mols C) = 1.7 x 1/2 = about 0.85

mols SiC formed if you had 1 mol SiO2 and all of C needed. That's 1 x (1 mol SiC/1 mol SiO2) = 1 x 1/1 = about 1 mol SiC.
You note that the values for mols SiC don't agree which means one of them is not right. The correct value in LR problems is ALWAYS the smaller value ad the reagent producing that value is the LR. So C is the limiting reagent and you could produce approx 0.8 mols SiC.

Now convert 0.8g SiC to grams. g = mols x molar mass.