if 20g of CaSiO3 is mixed with 15g of HF which of the two is the limiting reactant and how many grams of the other will be left
To determine the limiting reactant and the amount of the other reactant left, we need to calculate the moles of each reactant and then compare them based on the balanced chemical equation for the reaction.
First, let's calculate the number of moles for each reactant:
1. Calcium Silicate (CaSiO3):
- The molar mass of CaSiO3 can be calculated by summing the atomic masses of its constituents:
Molar mass of Ca = 40.08 g/mol
Molar mass of Si = 28.09 g/mol
Molar mass of O = 16.00 g/mol (x3 since there are three oxygen atoms in CaSiO3)
- Molar mass of CaSiO3 = (40.08 + 28.09 + (16.00 × 3)) g/mol = 116.16 g/mol
- To calculate the number of moles of CaSiO3, we will divide its mass by its molar mass:
Moles of CaSiO3 = mass of CaSiO3 / molar mass of CaSiO3
Moles of CaSiO3 = 20 g / 116.16 g/mol
Moles of CaSiO3 ≈ 0.172 mol
2. Hydrofluoric Acid (HF):
- The molar mass of HF can be calculated by summing the atomic masses of its constituents:
Molar mass of H = 1.01 g/mol
Molar mass of F = 19.00 g/mol
- Molar mass of HF = 1.01 + 19.00 = 20.01 g/mol
- To calculate the number of moles of HF, we will divide its mass by its molar mass:
Moles of HF = mass of HF / molar mass of HF
Moles of HF = 15 g / 20.01 g/mol
Moles of HF ≈ 0.749 mol
Now, let's look at the balanced chemical equation for the reaction between CaSiO3 and HF:
CaSiO3 + 6HF → CaF2 + SiF4 + 3H2O
According to the balanced equation, the stoichiometric ratio between CaSiO3 and HF is 1:6. This ratio indicates that 1 mole of CaSiO3 requires 6 moles of HF.
Next, we need to determine the limiting reactant by comparing the number of moles of CaSiO3 and HF. Since we have 0.172 moles of CaSiO3 and 0.749 moles of HF, we can use the stoichiometric ratio to calculate the number of moles required for a complete reaction.
Using the stoichiometric ratio of 1:6 (CaSiO3: HF), we can calculate the maximum number of moles of CaSiO3 that can react with the available moles of HF:
Max moles of CaSiO3 that can react = moles of HF / stoichiometric ratio
Max moles of CaSiO3 that can react = 0.749 mol / 6
Max moles of CaSiO3 that can react ≈ 0.125 mol
Since we have fewer moles of CaSiO3 (0.172 mol) than the maximum moles that can react (0.125 mol), CaSiO3 is the limiting reactant in this reaction.
To calculate the amount of the other reactant (HF) that will be left unused, we can subtract the moles of HF required for the complete reaction from the available moles of HF:
Moles of HF left = moles of HF - (moles of CaSiO3 * stoichiometric ratio)
Moles of HF left = 0.749 mol - (0.172 mol * 6)
Moles of HF left ≈ 0.749 mol - 1.032 mol
Moles of HF left ≈ -0.283 mol
The negative value indicates that there is not enough CaSiO3 for a complete reaction, so no HF will be left over.
Therefore, in this scenario, CaSiO3 is the limiting reactant, and no HF will be left unused.