Stoichiometry Reaction

Question

For the reaction of 0.1575 moles of magnesium bromide with an excess of silver nitrate according to the given equation:

MgBr2 (aq) + AgNO3 (aq) → AgBr (s) + Mg(NO3)2 (aq)

#1.) Calculate how many grams of AgBr can be produced.

#2) How much silver nitrate (in grams) is will be required to consume all the limiting reagent?

#3.) What would be the actual amount of AgBr produced if the reaction is known to only have 65% yield?

#4.) How much Mg(NO3)2 will be produced (actual) if the reaction yield of 65% is applied?

Answer 1

So, you posted this question yesterday, and apparently got no answers. Part of the reason for that is that you've posted several questions of a similar type, and the tutors get tired of doing homework for people, especially when those people don't even ATTEMPT to do the problem. So, for this problem, I will get you started, give you suggestions, and then if you still can't get the answer, you can post again with what you attempted, and what you don't understand.

First, always start with a correctly balanced equation. The equation given in the problem IS NOT balanced.

MgBr2(aq) + 2AgNO3(aq) ==> 2AgBr(s) + Mg(NO3)2(aq) .. BALANCED EQUATION

Since there is excess AgNO3, you don't have to find the limiting reactant. It will be MgBr2

(1). Use the balanced equation to find moles of AgBr formed from 0.1575 mols MgBr2. Then convert to grams of AgBr using molar mass of AgBr

(2). Use the balanced equation to find moles of AgNO3 will be used to react with 0.1575 mols MgBr2. Convert that to grams AgNO3 using the molar mass of AgNO3

(3). Divide the answer in (1) by 0.65

(4). Use the balanced equation to find moles of Mg(NO3)2 formed from 0.1575 mols MgBr2. Convert this to grams of Mg(NO3)2 using the molar mass of Mg(NO3)2. Multiply this value by 0.65.

Good luck.