Question

Consider the reaction: 3 Br2(l) + 8 NH3(g)→ 6 NH4Br(s) + N2(g)A) How many moles of ammonia are required to react with 0.60 mol of elemental bromine?

How many moles of ammonium bromide will be formed when 0.32 mol of ammonia react with excess elemental bromine?

What mass of chlorine gas is required to react with 25.0 g of carbon disulfide gas? CS2(g) + 3Cl2(g) S2Cl2(g) + CCl4(g)

Answer 1

3 mole of bromine react with =8 mole of NH3

.6 mole of bromine react to= 8×.6/3 mole NH3

= 1.6 mole NH3

8 mole of NH3 produce =6 mole of NH4Br

.32 mole of NH3 produce= 6×.32/8 mole NH4Br

=.24 mole NH4Br

No. Of mole in 25 g of CS2 = mass/ molar mass

= 25g / 76.14 g mole-1= .328 mole CS2

1 mole of CS2 require 3 mole of Cl2

So .328 mole CS2 require = 3× .328 mole Cl2

= .985 mole Cl2

Mass of Cl2 = mole × molar mass of Cl2

= .985 mole × 70.9 g mole-1

= 69.84 g Cl2