Question

35.5 g of silver nitrite is reacted with 35.5 grams of sodium sulfide which produces silver sulfide and sodium nitrite. Write and balance the equation and calculate the number of grams of silver sulfide produced.

A) AgNO₃ + Na₂S → Ag₂S + 2NaNO₂; 71.0 g
B) 2AgNO₂ + Na₂S → Ag₂S + 2NaNO₃; 35.5 g
C) AgNO₃ + 2NaS → Ag₂S + 2NaNO₃; 35.5 g
D) AgNO₃ + 2NaS → Ag₂S + 2NaNO₂; 71.0 g

Answer 1

The balanced equation for the reaction is AgNO₃ + 2NaS → Ag₂S + 2NaNO₂. The number of grams of silver sulfide produced is 71.0 g.

Step-by-step explanation:

The balanced equation for the reaction between silver nitrate (AgNO₃) and sodium sulfide (Na₂S) to produce silver sulfide (Ag₂S) and sodium nitrite (NaNO₂) is:

AgNO₃ + 2NaS → Ag₂S + 2NaNO₂

To calculate the number of grams of silver sulfide produced, we need to determine the limiting reactant and use stoichiometry. Given that 35.5 grams of silver nitrite and 35.5 grams of sodium sulfide are reactants, we assume they are both the limiting reactant since their masses are equal. Using the molar mass, we find that the molar mass of silver sulfide is 247.8 g/mol. Therefore, the number of grams of silver sulfide produced is 71.0 g.