Question

If 8.500 g of sodium nitrate and 7.500 g of potassium chloride are dissolved in 25.0 mL of water and heated to 80°C no solid Watermain. Using these amounts of potassium chloride and sodium nitrate, calculate the theoretical number of grams of sodium chloride and potassium nitrate that could be formed.

Answer 1

To calculate the theoretical number of grams of sodium chloride and potassium nitrate that could be formed, we need to determine the limiting reactant and use stoichiometry. The limiting reactant is potassium chloride, and we can form 73.0 grams of both sodium chloride and potassium nitrate from the given amounts of reactants.

Step-by-step explanation:

To calculate the theoretical number of grams of sodium chloride and potassium nitrate that could be formed, we need to determine the limiting reactant and use stoichiometry. The balanced equation for the reaction between sodium nitrate (NaNO3) and potassium chloride (KCl) is:

NaNO3 + KCl -> NaCl + KNO3

First, we need to calculate the number of moles of sodium nitrate and potassium chloride using their molar masses (85.0 g/mol for NaNO3 and 74.55 g/mol for KCl). By dividing the given masses by their respective molar masses, we find that we have 0.100 mol of NaNO3 and 0.101 mol of KCl.

Next, we determine the limiting reactant by comparing the mole ratios of NaNO3 and KCl in the balanced equation. The ratio is 1:1, and since we have more moles of KCl (0.101 mol) than NaNO3 (0.100 mol), KCl is the limiting reactant.

Finally, we use stoichiometry to calculate the theoretical grams of products. From the balanced equation, we know that 1 mole of KCl yields 1 mole of NaCl. Therefore, the theoretical number of grams of sodium chloride formed is 73.0 g (the molar mass of NaCl is 58.44 g/mol).

Similarly, 1 mole of KCl yields 1 mole of KNO3. So, the theoretical number of grams of potassium nitrate formed is also 73.0 g (the molar mass of KNO3 is 101.10 g/mol).

Answer 2

Zero grams of NaCl & KNO₃ ... The reagents mixed in aqueous solution would not produce products as no driving force is present; i.e., no precipitate, weak electrolyte or gas decomposition product will form.

Step-by-step explanation:

NaNO₃(aq) + KCl(aq) => (theoritically) NaCl(aq) + KNO₃(aq)

However all chloride salts from metathesis rxns remain soluble and 100% ionized except for halide derivatives of Hg⁺², Ag⁺ & Pb⁺₂. Further, all NO₃⁻ salt derivatives also remain soluble and 100% ionized.

Technically, metathesis reactions (also known as double replacement rxns) depend upon the formation of a 'driving force' that leaves the reaction system in the form of a precipitate, weak electrolyte (weak acid or weak base) or a gas decomposition product. So, the reaction in this case should be as follows:

NaNO₃(aq) + KCl(aq) => No Reaction