To solve this problem, we first need to write the balanced chemical equation for the reaction between silver oxide and hydrochloric acid. The balanced chemical equation is:
2Ag2O + 4HCl -> 4AgCl + 2H2O
From the balanced chemical equation, we can see that 2 moles of silver oxide react with 4 moles of hydrochloric acid to produce 4 moles of silver chloride and 2 moles of water.
Next, we can use this information to calculate the amount of silver oxide needed to react with 7.9 g of hydrochloric acid.
First, we calculate the number of moles of hydrochloric acid (HCl) using its molar mass:
Molar mass of HCl = 1.007 grams/mole (for H) + 35.453 grams/mole (for Cl) = 36.46 grams/mole
Number of moles of HCl = mass of HCl / molar mass of HCl
= 7.9 g / 36.46 g/mol
≈ 0.216 moles of HCl
From the balanced chemical equation, we know that 2 moles of Ag2O react with 4 moles of HCl. Therefore, we can use the mole ratio to calculate the number of moles of Ag2O needed:
Moles of Ag2O = (0.216 moles of HCl / 4 moles of HCl) * 2 moles of Ag2O
≈ 0.108 moles of Ag2O
Finally, we can convert the moles of Ag2O to grams:
Molar mass of Ag2O = 2*(107.87 g/mol) + 16.00 g/mol
= 231.74 g/mol
Mass of Ag2O = moles of Ag2O * molar mass of Ag2O
≈ 0.108 moles * 231.74 g/mol
≈ 25.04 grams
Therefore, approximately 25.04 grams of silver oxide are needed to react with 7.9 g of hydrochloric acid.
