When silver nitrate reacts with barium chloride,silver chloride and barium nitrate are formed.How many grams of silver chloride are formed when 8.0 g of silver nitrate reacts with 15.0 g of barium chloride?

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Quizplus · Accepted Answer

First, we need to write the balanced chemical equation for the reaction: $ \text{2AgNO}_3 + \text{BaCl}_2 \rightarrow 2\text{AgCl} + \text{Ba(NO}_3)_2 $. Then, calculate the molar mass of silver nitrate ($ \text{AgNO}_3 $) and barium chloride ($ \text{BaCl}_2 $). The molar mass of $ \text{AgNO}_3 $ is approximately 169.87 g/mol, and for $ \text{BaCl}_2 $, it's about 208.23 g/mol. The molar mass of silver chloride ($ \text{AgCl} $) is approximately 143.32 g/mol. Using stoichiometry, we find that 8.0 g of $ \text{AgNO}_3 $ (which is 8.0 g / 169.87 g/mol = 0.0471 mol) can theoretically produce 0.0471 mol × 2 = 0.0942 mol of $ \text{AgCl} $, because the reaction ratio is 1:2. However, we must also check if $ \text{BaCl}_2 $ is in excess. For $ \text{BaCl}_2 $, 15.0 g / 208.23 g/mol = 0.072 mol, which is more than enough to react with the 0.0471 mol of $ \text{AgNO}_3 $. Therefore, $ \text{AgNO}_3 $ is the limiting reagent. The mass of $ \text{AgCl} $ produced is 0.0942 mol × 143.32 g/mol = 13.5 g, but due to the stoichiometry of the reaction (2:1 ratio of $ \text{AgNO}_3 $ to $ \text{AgCl} $), the correct calculation should consider the 1:1 mole ratio in the product side, leading to 0.0471 mol of $ \text{AgNO}_3 $ producing 0.0942 mol of $ \text{AgCl} $, which gives approximately 6.752 g of $ \text{AgCl} $ when calculated correctly. My initial explanation contained a miscalculation regarding the stoichiometry and the final mass of $ \text{AgCl} $ produced. The correct approach is directly using the limiting reagent's moles to find the moles of $ \text{AgCl} $ produced and then converting those moles to grams.

When Silver Nitrate Reacts with Barium Chloride,silver Chloride and Barium