Question

Na2SiO3 (s) + 8HF(aq) → H2SiF6 (aq) + 2NaF(aq) + 3H2O (l)

How many moles of HF are needed to react with 87.44 mol of Na2SiO3?
How many grams of NaF form when 0.94 mol of HF reacts with excess Na2SiO3?
How many liters of Na2SiO3 can react with 54.68 g of HF?
a) 699.52 mol, 61.72 g, 17.84 L
b) 174.88 mol, 49.28 g, 35.68 L
c) 87.44 mol, 98.56 g, 8.92 L
d) 349.76 mol, 74.96 g, 26.78 L

Answer 1

To react with 87.44 moles of Na2SiO3, 699.52 moles of HF are required. When 0.94 moles of HF react, 9.87 g of NaF are formed. Lastly, 54.68 g of HF would react with 0.342 L of Na2SiO3, assuming a 1M concentration.

Step-by-step explanation:

To answer the question involving the reaction of Na2SiO3 with HF, we must first use the stoichiometry of the provided chemical equation. According to the balanced equation, 8 moles of HF are required to react with 1 mole of Na2SiO3. Therefore:

1. To react with 87.44 moles of Na2SiO3, we need 87.44 moles × 8 moles of HF/1 mole of Na2SiO3 = 699.52 moles of HF.
2. To find out how many grams of NaF form when 0.94 moles of HF react, we first determine the moles of NaF that would form. Since the ratio of HF to NaF is 8:2 (or 4:1), 0.94 moles of HF would produce 0.94/4 = 0.235 moles of NaF. The molar mass of NaF is 41.99 g/mol, so 0.235 moles × 41.99 g/mol = 9.86765 g (rounded to 9.87 g) of NaF.
3. To calculate how many liters of Na2SiO3 can react with 54.68 g of HF, we first convert the mass of HF to moles (molar mass of HF being approximately 20 g/mol), which is 54.68 g / 20 g/mol = 2.734 moles of HF. Using the stoichiometry of the reaction (8:1), we find that 2.734 moles of HF would react with 2.734/8 = 0.34175 moles of Na2SiO3. If the molarity of Na2SiO3 is 1 M, then 0.34175 moles would correspond to 0.34175 liters (rounded to 0.342 L) of Na2SiO3 at 1 M concentration.

Thus, the answers are 699.52 moles of HF, 9.87 g of NaF, and 0.342 L of Na2SiO3 when using the correct stoichiometry and assumptions about concentrations.