A 5.98-g sample consisting of a mixture of silver nitrate and sodium nitrate is dissolved in water. This aqueous mixture then reacts with aqueous barium chloride to form 3.90 g of silver chloride. Calculate the percent by mass of silver nitrate in the first mixture.
A) 77.3%
B) 55.0%
C) 46.2%
D) 53.2%
E) 94.7%

Question

Quizplus · Accepted Answer

The mass of silver chloride formed (3.90 g) can be used to calculate the mass of silver nitrate in the original mixture. The molar mass of AgCl is 143.32 g/mol, and the molar mass of AgNO3 is 169.87 g/mol. Using stoichiometry, the mass of AgNO3 that reacted to form 3.90 g of AgCl can be calculated. Since the reaction is 1:1 for AgNO3 to AgCl, the moles of AgCl formed are equal to the moles of AgNO3 that reacted. $$ \text{Moles of AgCl} = \frac{3.90 \, \text{g}}{143.32 \, \text{g/mol}} = 0.0272 \, \text{mol} $$Since the moles of AgNO3 are equal to the moles of AgCl, the mass of AgNO3 in the mixture is:$$ \text{Mass of AgNO3} = 0.0272 \, \text{mol} \times 169.87 \, \text{g/mol} = 4.62 \, \text{g} $$The percent by mass of AgNO3 in the mixture is then:$$ \text{Percent by mass of AgNO3} = \left( \frac{4.62 \, \text{g}}{5.98 \, \text{g}} \right) \times 100\% = 77.3\% $$

A 598-G Sample Consisting of a Mixture of Silver Nitrate and and Sodium