For the following reaction, what mass of SIC is made from120.2g of SIO2 and 120g of C?

SiO2(s) + 3C(s) ---SiC(s) +2CO(g)

This post is a limiting reagent problem. I know because values for BOTH reactants are given. Here is a example of a simple stoichiometry problem. Just follow the instructions for EACH reactant. The correct one will be the one providing the smaller value for the product and that reagent is the limiting reagent.

http://www.jiskha.com/science/chemistry/stoichiometry.html

If 100 grams of SiO2 and 120 grams of C react, Calculate the moles of both of them

To find the mass of SiC produced, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and limits the amount of product formed.

Step 1: Calculate the molar masses of the reactants and the product.
Molar mass of SiO2 = 60.08 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of SiC = 40.10 g/mol

Step 2: Calculate the number of moles for each reactant.
Number of moles of SiO2 = mass / molar mass = 120.2 g / 60.08 g/mol = 2.001 mol
Number of moles of C = mass / molar mass = 120 g / 12.01 g/mol = 9.991 mol

Step 3: Determine the mole ratio between the reactants and the product.
From the balanced chemical equation, the mole ratio between SiO2 and SiC is 1:1. Therefore, the moles of SiO2 and SiC are equal.

Step 4: Using the mole ratio, identify the limiting reactant.
Since the moles of SiO2 and SiC are equal, we compare the moles of C with the mole ratio, which is 3:1. With 9.991 mol of C, we can only produce a maximum of 9.991 / 3 = 3.33 mol of SiC. However, we have 2.001 mol of SiO2. Therefore, C is the limiting reactant.

Step 5: Calculate the mass of SiC produced.
To determine the mass of SiC produced, we multiply the number of moles of SiC by its molar mass:
Mass of SiC = moles of SiC × molar mass of SiC = 2.001 mol × 40.10 g/mol = 80.18 g.

Therefore, the mass of SiC produced from 120.2g of SiO2 and 120g of C is 80.18 g.

To find the mass of SiC produced, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.

To find the limiting reactant, we can compare the amounts of SiO2 and C and see which one is present in a lower amount relative to their stoichiometric ratio in the balanced equation.

First, we need to calculate the number of moles of SiO2 and C:

1) Moles of SiO2 = mass of SiO2 / molar mass of SiO2
= 120.2 g / 60.08 g/mol (molar mass of SiO2)
≈ 2.00 mol

2) Moles of C = mass of C / molar mass of C
= 120 g / 12.01 g/mol (molar mass of C)
≈ 10.00 mol

Now, let's calculate the stoichiometric ratios for SiO2 and C:

Stoichiometric ratio for SiO2:SiC = 1:1
Stoichiometric ratio for C:SiC = 3:1

Now, we can consider the reaction with SiO2 as the limiting reactant:

Moles of SiC produced from SiO2 = 2.00 mol × (1 mol SiC / 1 mol SiO2)
= 2.00 mol

Now, let's consider the reaction with C as the limiting reactant:

Moles of SiC produced from C = 10.00 mol × (1 mol SiC / 3 mol C)
≈ 3.33 mol

Comparing the number of moles of SiC produced, 2.00 mol (from SiO2) is less than 3.33 mol (from C). Therefore, SiO2 is the limiting reactant.

Next, we can calculate the mass of SiC produced from the moles of SiC:

Mass of SiC = moles of SiC × molar mass of SiC
= 2.00 mol × 40.10 g/mol (molar mass of SiC)
= 80.20 g

Therefore, the mass of SiC produced from 120.2 g of SiO2 and 120 g of C is 80.20 g.