Iodine monobromide and ammonia react to produce nitrogen triiodide and ammonium bromide.
(a) Balance the chemical equation for this reaction. (Use the smallest possible whole numbers. Enter 1 where appropriate. Do not leave any answer box blank.)
. IBr(s) + . NH3(g) . NI3(s) + . NH4Br(s)
(b) How many moles of ammonium bromide could be produced if you start with 0.750 moles of ammonia and excess iodine monobromide?
mol
To balance the chemical equation, we need to ensure that the number of each type of atom is the same on both sides of the equation.
(a) The balanced chemical equation is:
1 IBr(s) + 3 NH3(g) -> 1 NI3(s) + 3 NH4Br(s)
(b) To find the number of moles of ammonium bromide produced, we need to use the stoichiometry of the balanced equation. From the equation, we can see that 1 mole of IBr reacts with 3 moles of NH3 to produce 3 moles of NH4Br.
Given that we have 0.750 moles of NH3, we can set up a stoichiometric ratio:
(0.750 moles NH3) * (3 moles NH4Br / 3 moles NH3) = 0.750 moles of NH4Br
Therefore, 0.750 moles of ammonium bromide could be produced.
To balance the chemical equation for the reaction between iodine monobromide (IBr) and ammonia (NH3), we need to ensure that the number of moles of each element is equal on both sides of the equation.
(a) The balanced chemical equation is:
1 IBr(s) + 3 NH3(g) → 1 NI3(s) + 3 NH4Br(s)
(b) To find the number of moles of ammonium bromide (NH4Br) that can be produced from 0.750 moles of ammonia (NH3), we need to use the mole ratio between ammonia and ammonium bromide.
From the balanced equation, we can see that 3 moles of NH4Br are produced from 3 moles of NH3. Therefore, the mole ratio is 1:1.
Therefore, if you start with 0.750 moles of ammonia, you would produce 0.750 moles of ammonium bromide.
For b just follow the steps is this example problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
For the equation it makes it easier if you show the arrow. Otherwise we must guess.
3IBr + 4NH3 ==>NI3 + 3NH4Br
A couple of notes. It is true that NH3 is a gas but little reaction will occur between gas NH3 and the solid IBr. Usually this is carried out in an aqueous solution, which dissolves NH3 readily, and the NH4Br then is in aqueous solution.