When silver nitrate reacts with barium chloride,silver chloride and barium nitrate are formed.How many grams of silver chloride are formed when 10.0 g of silver nitrate reacts with 15.0 g of barium chloride?
A)8.44 g
B)9.40 g
C)11.9 g
D)18.8 g

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction between silver nitrate and barium chloride is:

AgNO3 + BaCl2 -> AgCl + Ba(NO3)2

To determine the amount of silver chloride formed, we first need to identify the limiting reactant. This can be done by calculating the amount of each reactant needed to completely react with each other based on their stoichiometric ratio.

From the balanced equation, we can see that:

1 mole of AgNO3 reacts with 1 mole of BaCl2 to produce 1 mole of AgCl

The molar masses of each compound are:

AgNO3 = 169.9 g/mol
BaCl2 = 208.3 g/mol

Using these values, we can calculate the number of moles of each reactant:

Number of moles of AgNO3 = 10.0 g / 169.9 g/mol = 0.05886 mol
Number of moles of BaCl2 = 15.0 g / 208.3 g/mol = 0.07202 mol

Based on the stoichiometric ratio, we can see that AgNO3 is the limiting reactant since it produces less moles of AgCl compared to BaCl2. Therefore, we will use the amount of AgNO3 to calculate the amount of AgCl formed.

Number of moles of AgCl = 0.05886 mol

The molar mass of AgCl is:

AgCl = 143.3 g/mol

Using this value, we can calculate the mass of AgCl formed:

Mass of AgCl = Number of moles of AgCl x Molar mass of AgCl
Mass of AgCl = 0.05886 mol x 143.3 g/mol
Mass of AgCl = 8.44 g

Therefore, the answer is (A) 8.44 g.

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