Answer to Question #123706 in General Chemistry for Hamda jumaa

Question

3.38 g of AgNO3 react with BaCl2 to produce 1.43 g of AgCl according to the following balanced equation: 2AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2 What is the percent yield of AgCl in this reaction?

Answer 1

According to the equation, moles AgNO₃ = moles AgCl.

So the theoretical yield of this reaction can be calculated as:

m_theor(AgCl) = (m(AgNO₃) / M(AgNO₃)) x M (AgCl) = (3.38g / 169.87g/mol) x 143.32g/mol = 2.85 g.

%yield = (m_actual / m_theor) x 100% = (1.43g / 2.85g) x 100% = 50.2%

Answer: 50.2%