Predict the mass of gold formed when a gold (III) nitrate solution reacts completely with 125 mL of 0.352 mol/L sulfurous acid (aqueous hydrogen sulfite)
To predict the mass of gold formed when a gold (III) nitrate solution reacts completely with sulfurous acid, we need to balance the equation and calculate the stoichiometry of the reaction.
The balanced chemical equation for the reaction between gold (III) nitrate and sulfurous acid is as follows:
Au(NO3)3(aq) + 3 H2SO3(aq) → Au(s) + 3 HNO3(aq) + 3 H2O(l)
From the balanced equation, we can see that 1 mole of Au(NO3)3 reacts with 3 moles of H2SO3 to produce 1 mole of Au. Therefore, the stoichiometric ratio of Au(NO3)3 to Au is 1:1.
To calculate the amount of gold formed, we first need to determine the number of moles of H2SO3 in the solution.
Given:
Volume of sulfurous acid solution (V) = 125 mL = 125/1000 L = 0.125 L
Concentration of sulfurous acid solution (C) = 0.352 mol/L
Using the equation: moles = concentration × volume
moles of H2SO3 = C × V = 0.352 mol/L × 0.125 L = 0.044 mol
Since the stoichiometry between H2SO3 and Au is 3:1, we can calculate the number of moles of Au produced.
moles of Au = 0.044 mol/3 = 0.0147 mol
Now, we need to find the molar mass of gold (Au) from the periodic table, which is approximately 197 g/mol.
To determine the mass of gold formed, we can use the equation: mass = moles × molar mass
mass of gold = 0.0147 mol × 197 g/mol = 2.89 grams
Therefore, the predicted mass of gold formed when a gold (III) nitrate solution reacts completely with 125 mL of 0.352 mol/L sulfurous acid is approximately 2.89 grams.