Silicon carbide is produced by heating silicon dioxide and carbon at high temperatures as shown by the reaction below. How many grams of SiC can be formed by reacting 2.00g of silicon dioxide and 2.00 grams of carbon?
SiO2(s) + 3 C(s) --> SiC(s) + 2 CO(g)
How much excess reactant is left over?
(show your work)
This is a limiting reagent (LR) problem. You know that when an amount is given for more than one reactant. I do these the long way.
SiO2(s) + 3 C(s) --> SiC(s) + 2 CO(g)
1a. mols SiO2 = grams/molar mass = 2.00/60.1 = 0.0333
1b. mols C = 2.00/12.01 = 0.166
2. How much SiC can be formed from each with an excess of the other?
2a. mols SiC=mols SiO2 x (1 mol SiC/1 mol SiO2) = 0.0333 x 1/1 = 0.0333
2b. mols SiC = mols C x (1 mol SiC/3 mols C) = 0.166 x 1/3 = 0.0553
2c. In LR problems the smaller number wins because no more than the smaller amount can be formed. So SiO2 is the LR and C is the excess reagent (ER).
2d, grams SiC = mols SiC x molar mass SiC = ?
3. ER
How much C is used forming that 0.0333 mols SiC.
mol C used = mols SiO2 used x (1 mol SiO2/3 mols C) = 0.0333 x 1/3 = 0.0111. You had 0.0333 initiall - 0.0111 used = 0.0222 mols C left.
If you want grams C remaining that is mols C remaining x atomic mass = ?
To find out how many grams of SiC can be formed and how much excess reactant is left over, we need to follow a few steps:
Step 1: Calculate the moles of each reactant.
To do this, we use the formula:
number of moles = mass (in grams) / molar mass (in grams/mol)
The molar mass of SiO2 is 60.08 g/mol, and the molar mass of C is 12.01 g/mol.
For silicon dioxide (SiO2):
Number of moles = 2.00 g / 60.08 g/mol = 0.0333 mol (rounded to 4 decimal places)
For carbon (C):
Number of moles = 2.00 g / 12.01 g/mol = 0.166 mol (rounded to 3 decimal places)
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the mole ratios of the reactants in the balanced equation.
From the balanced equation:
1 mol of SiO2 reacts with 3 mol of C to produce 1 mol of SiC.
The reactant with the smaller number of moles is the limiting reactant. In this case, SiO2 has the smaller number of moles.
Step 3: Calculate the moles of SiC formed.
From the mole ratio in the balanced equation, we know that:
1 mol of SiO2 reacts to produce 1 mol of SiC.
Therefore, the moles of SiC formed will be equal to the moles of SiO2 present, which is 0.0333 mol.
Step 4: Calculate the mass of SiC formed.
To calculate the mass of SiC, we use the formula:
mass (in grams) = number of moles × molar mass (in grams/mol)
The molar mass of SiC is 40.10 g/mol.
Mass of SiC formed = 0.0333 mol × 40.10 g/mol = 1.336 g (rounded to 3 decimal places)
Therefore, 1.336 grams of SiC can be formed.
Step 5: Determine the excess reactant.
To find the amount of excess reactant, we subtract the moles of the limiting reactant used from the total moles of the excess reactant.
In this case, the limiting reactant is SiO2. From step 1, we calculated that there are 0.0333 mol of SiO2. Since the mole ratio is 1:3 (SiO2:C), it means that 0.0333 mol of SiO2 reacts with 0.0333 mol × 3 = 0.0999 mol of C.
The total moles of C used are 0.0999 mol. From step 1, we calculated that there are 0.166 mol of C.
Therefore, the excess reactant is:
0.166 mol C - 0.0999 mol C = 0.0661 mol C
Step 6: Calculate the mass of the excess reactant.
To calculate the mass of the excess reactant, we use the same formula as in step 4, using the molar mass of C.
Mass of excess C = 0.0661 mol × 12.01 g/mol ≈ 0.793 g (rounded to 3 decimal places)
Therefore, approximately 0.793 grams of carbon are left over as excess reactant.