6HF + SiO2 → H2SiF6 + 2H2O
Identify the limiting reactant and excess reactant if 45 moles of SiO2 and 140 moles of HF react. (calculate grams of H2O)
well, you will need 6*45 = 270 moles of HF for that much SiO2.
To determine the limiting reactant and excess reactant, we need to compare the amount of reactants to the stoichiometric ratio in the balanced chemical equation.
First, let's calculate the number of moles of H2O that can be produced from each reactant:
From the balanced equation, we can see that 1 mole of SiO2 reacts to form 2 moles of H2O.
Therefore, the number of moles of H2O that can be produced from 45 moles of SiO2 is:
45 moles SiO2 * (2 moles H2O / 1 mole SiO2) = 90 moles H2O
Similarly, 1 mole of HF reacts to form 2 moles of H2O.
Therefore, the number of moles of H2O that can be produced from 140 moles of HF is:
140 moles HF * (2 moles H2O / 1 mole HF) = 280 moles H2O
Next, we compare the moles of H2O formed from each reactant to determine the limiting reactant. The reactant that produces a smaller amount of product is the limiting reactant.
For SiO2, the moles of H2O produced is 90 moles.
For HF, the moles of H2O produced is 280 moles.
Since SiO2 produces a smaller amount of H2O, it is the limiting reactant. HF is the excess reactant.
To calculate the grams of H2O, we need to use the molar mass of water (H2O), which is approximately 18.015 grams/mol.
The number of grams of H2O formed can be calculated using the limiting reactant (SiO2):
90 moles H2O * (18.015 grams H2O / 1 mole H2O) = 1622.35 grams H2O
Therefore, the number of grams of H2O formed is approximately 1622.35 grams.
To identify the limiting reactant and excess reactant in this reaction, we need to compare the given amounts of the reactants to their stoichiometric coefficients. The stoichiometric coefficients tell us the balanced ratio in which the reactants combine.
Let's calculate the number of moles of products that can be formed from the given amounts of reactants:
From the balanced equation, we can see that 1 mole of SiO2 reacts with 6 moles of HF to produce 1 mole of H2SiF6 and 2 moles of H2O.
Moles of H2SiF6 = 45 moles of SiO2 * (1 mole of H2SiF6 / 6 moles of SiO2) = 7.5 moles of H2SiF6
Moles of H2O = 45 moles of SiO2 * (2 moles of H2O / 6 moles of SiO2) = 15 moles of H2O
Now, let's compare the calculated moles of products to the given amount of HF:
Moles of HF = 140 moles of HF
From the balanced equation, we can see that 6 moles of HF react with 1 mole of SiO2 to produce 2 moles of H2O. Therefore, the moles of H2O that can be produced from the given moles of HF are given by:
Moles of H2O from HF = 140 moles of HF * (2 moles of H2O / 6 moles of HF) = 46.67 moles of H2O
From the calculations, we can determine that the limiting reactant is SiO2 because it produces a smaller amount of moles of H2O compared to HF. Therefore, HF is the excess reactant.
To calculate the grams of H2O produced, we need to use the molar mass of water (H2O), which is approximately 18.015 g/mol.
Grams of H2O = Moles of H2O * Molar mass of H2O
Grams of H2O = 15 moles of H2O * 18.015 g/mol ≈ 270.23 g
Therefore, approximately 270.23 grams of H2O can be produced from the given amounts of reactants.