When silver nitrate reacts with barium chloride,silver chloride and barium nitrate are formed.How many grams of silver chloride are formed when 10.2 g of silver nitrate reacts with 15.0 g of barium chloride?
A) 8.61 g
B) 9.40 g
C) 12.1 g
D) 18.8 g

Question

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 $.Next, we calculate the molar masses of the reactants and products involved in the reaction:- Molar mass of $ \text{AgNO}_3 $ = 169.87 g/mol- Molar mass of $ \text{BaCl}_2 $ = 208.23 g/mol- Molar mass of $ \text{AgCl} $ = 143.32 g/molUsing the given masses of the reactants, we find the moles of each reactant:- Moles of $ \text{AgNO}_3 $ = $ \frac{10.2 \text{ g}}{169.87 \text{ g/mol}} $ = 0.0600 mol- Moles of $ \text{BaCl}_2 $ = $ \frac{15.0 \text{ g}}{208.23 \text{ g/mol}} $ = 0.0720 molFrom the balanced equation, we see that 2 moles of $ \text{AgNO}_3 $ react with 1 mole of $ \text{BaCl}_2 $. Therefore, $ \text{AgNO}_3 $ is the limiting reactant in this case.The reaction produces 2 moles of $ \text{AgCl} $ for every 2 moles of $ \text{AgNO}_3 $ reacted. So, the moles of $ \text{AgCl} $ produced are equal to the moles of $ \text{AgNO}_3 $ reacted, which is 0.0600 mol.Finally, we calculate the mass of $ \text{AgCl} $ produced:- Mass of $ \text{AgCl} $ = $ 0.0600 \text{ mol} \times 143.32 \text{ g/mol} $ = 8.60 g, which rounds to 8.61 g when considering significant figures.

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