Question

Will give brainliest if you show work and have the right answer :)

Sulfur dioxide reacts with bromine and water to produce hydrogen bromide and sulfuric acid. If 64.1 g of sulfur dioxide react completely with 159.9 g of bromine and an unknown amount of water to form 161.9 g of hydrogen bromide and 98.1 g of sulfuric acid, then how many grams of water react? (include units)

Answer 1

Answer:

The answer to your question is:

a) 36 g of water

Explanation:

MW SO₂ = 32 + 32 = 64 g

MW Br = 80 g

MW H₂O = 18 g

MW HBr = 81 g

MW H₂SO₄ = 98

Balanced reaction:

                      SO₂   +   2Br   +   2H₂O    ⇒    2HBr   +   H₂SO₄

                     64g       2(80)        2(18)             2(81)            98

                     64.1       159.9                           161.9             98.1

Rule of three with Br and H2O

                              160 g ----------------   36 g

                              160 -------------------    x

           x = 160(36)/160 = 36 g oh water

                   

Answer 2

Answer: The mass of water reacted is 36 grams

Explanation:

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the reaction of sulfur dioxide with bromine and water follows:

$SO_2+Br_2+2H_2O\rightarrow 2HBr+H_2SO_4$

Let the mass of water be 'x' grams

We are given:

Mass of sulfur dioxide = 64.1 grams

Mass of bromine = 159.9 grams

Mass of hydrogen bromide = 161.9 grams

Mass of sulfuric acid = 98.1 grams

Total mass on reactant side = 64.1 + 159.9 + x  = 224 + x

Total mass on product side = 161.9 + 98.1 = 260 g

So, by applying law of conservation of mass, we get:

$224+x=260\\\\x=(260-224)=36g$

Hence, the mass of water reacted is 36 grams.