Question

What quantity in moles of precipitate will be formed when 84.3 ml of 0.250 M AgNO₃ is reacted with excess CaBr₂ in the following chemical reaction? 2 AgNO₃(aq) + CaBr₂(aq) → 2 AgBr(s) + Ca(NO₃)₂(aq)

Answer 1

The quantity of precipitate in moles formed when 84.3 ml of 0.250 M AgNO₃ is reacted with excess CaBr₂ is 0.250 moles.

Step-by-step explanation:

To find the quantity of precipitate in moles, we need to use the balanced chemical equation and the given information. The balanced equation is:

2 AgNO₃(aq) + CaBr₂(aq) → 2 AgBr(s) + Ca(NO₃)₂(aq)

From the equation, we can see that 2 moles of AgNO₃ react to form 2 moles of AgBr. Since the molar ratio is 1:1, this means that 84.3 ml of 0.250 M AgNO₃ will also produce 84.3 ml of AgBr.

Therefore, the quantity in moles of precipitate formed is the same as the quantity in moles of AgNO₃, which is

0.250 moles (since the concentration is given as 0.250 M).