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 = ?
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)
1 answer