Answered

Discover new information and get your questions answered with IDNLearn.com. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.

Excess silver(I) nitrate was added to a 8.500 g mixture containing some amount of barium chloride, and 7.123 g of silver chloride was obtained. The unbalanced equation is
AgNO3 + BaCl2(aq) --> Ba(NO3)2(aq) + AgCl(s).
What is the mass% of BaCl2 in the mixture?

Sagot :

Answer:

60.88%

Explanation:

The balanced equation of the reaction is given as;

2 AgNO3 (aq) + BaCl2 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)

Since AgNO3 is in excess, the limiting reactant is BaCl2. From the reaction;

1 mol of BaCl2 produces 2 mol of AgCl

Converting to masses;

Mass = Number of mol * Molar mass

BaCl2;

Mass = 1 * 208.23 g/mol = 208.23 g

AgCl;

Mass = 2 * 143.32 g/mol = 286.64 g

208.23 g BaCl2 produces 286.64 g of AgCl

x g BaCl2 produces 7.123 g of AgCl

Solving for x;

x = 7.123 * 208.23 / 286.64 = 5.1745 g

Mass percent = Mass / Total mass of Mixture  * 100

Mass Percent = 5.1745 / 8.500 = 0.6088 * 100 = 60.88%

We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.