When silicon dioxide (sand) and carbon are heated, the products are silicon carbide, SiC, and carbon monoxide, CO.Silicon carbide is a ceramic material that tolerates extreme temperatures and is used as an abrasive and in the brake discs of cars. Given 70.0 g of Si02 and 50.0 g of C, which is the limiting reactant?

To determine the limiting reactant, we need to calculate the amount of product that can be formed from each reactant using the stoichiometric coefficients from the balanced chemical equation.

SiO2 + 3C -> SiC + 2CO

Let's call the amount of SiO2 used in the reaction "x" and the amount of C used in the reaction "y". We can then write the following relationships:

x = 70.0 g SiO2
y = 50.0 g C

From the stoichiometric coefficients, we can also write:

x = (1 mole SiO2 / 60.1 g SiO2) * (1 mole SiC / 1 mole SiO2) * (40.1 g SiC / mole SiC)
y = (3 moles C / 12.0 g C) * (1 mole SiC / 3 moles C) * (40.1 g SiC / mole SiC)

We can equate the two expressions for SiC to determine which reactant is limiting:

(1 mole SiO2 / 60.1 g SiO2) * x = (3 moles C / 12.0 g C) * y

x / 60.1 = 3y / 12.0

Multiplying both sides by the denominators, we get:

x = 3y * (60.1 g SiO2 / 12.0 g C)

Now we can substitute the values of x and y to see which one is limiting:

70.0 g SiO2 = 3 * 50.0 g C * (60.1 g SiO2 / 12.0 g C)
70.0 g SiO2 = 75.0 g SiO2

Since 70.0 g SiO2 is less than 75.0 g SiO2, we can conclude that SiO2 (silicon dioxide) is the limiting reactant in this reaction.