Hydrogen gas and bromine gas react to form hydrogen bromide gas. Can anyone help?

Question

a)Write a balanced chemical equation for this reaction.

b)How many grams of hydrogen bromide gas can be produced from 3.2 g of hydrogen gas and 9.5 g of bromine gas?

c)How many grams of which reactant is left unreacted?

d)What volume of HBr, measured at STP, is produced in b?

Answer 1

a) Br2 + H2 => 2HBr

I wont awnser the rest as my chemistry days are long gone.

Answer 2

a)  H2 + Br2  ----->  2HBr

b)  The molar mass of H2 is 2.016 g/mol; the molar mass of Br2 is 159.8 g/mol.

3.2 g H2 x (1 mol / 2.016 g) = 1.59 mol H2

9.5 g Br2 x (1mol / 159.8 g) = 0.0594 mol Br2

Since 1 mol H2 reacts with 1 mol Br2, Br2 is the limiting reagent because there are fewer moles of Br2. Use Br2 to find the grams of HBr produced.

The molar mass of HBr is 80.91 g/mol.

1 mol Br2 produces 2 mol HBr

0.0594 mol Br2 x (2 mol HBr / 1 mol Br2) = 0.119 mol HBr

0.119 mol HBr x (80.91 g / 1 mol ) = 9.61 g HBr

c) Excess mol of H2 = 1.59 - 0.0594 = 1.53 mol H2

1.53 mol H2 x (2.016 g / 1 mol H2) = 3.09 g H2 left over

d) 1 mol of gas occupies 22.4 L at STP.

0.119 mol HBr x (22.4 L / 1 mol) = 2.67 L

Answer 3

a) H2 + Br2 = 2 HBr

Principle: you write correct formulas for all reactants and products, and you have to have equal numbers of every kind of atom on both sides because atoms are neither created or destroyed.

b) Divide the mass of H2 by the molar mass of H2 to get the number of moles of H2. Similarly, find the number of moles of Br2.

Then see which one you are going to run out of first.  This is what is called the "limiting reagent".

The maximum amount of moles of HBr that you can make is determined by the limiting reagent.  (actually, in this case, the equation tells you that you get 2 moles HBr for every mole limiting reagent). To get the maximum mass of HBr that you can make, multiply number of moles of HBr by molar mass of HBr.

c) Take away the number of moles of the other reagent (the one said to be "present in excess") that were used up in the reaction.  That is the number of moles you have left, so multiply that by molar mass to get number of grams you have left.

d) There are two ways of doing this.  You know n, the number of moles of HBr, because you found it on the way to solving (b). So you can either use the fact that one mole gas at STP occupies 22.4 L, or if you have already done the gas laws you can use the equation

PV = nRT

which is better because the equation works under any conditions.

Remember that this is just bookkeeping, an exercise in keeping track of the amounts of stuff, and in going from grams to moles and back again.

Good luck.