How many grams of bromine will be required to completely react with 10 g of propyne?

(C = 12, H = 1, Br = 80)

chydirim@gmail.com

10 years ago

This can't be the answer. When ethyne reacts completly with bromine(addition reaction), it doesn't produce hydrogen. The equation ought to be

C3H4+Br2-C3H4Br2

(36+4)(80*2)

40g of ethyne reacts with 160g of bromine

10g of ethyne reacts with xg of bromine

40/10=160/x

10*160=40*x

1600/40=x

x=40g. optionB

ELIJAH0016

1 year ago

The chemical equation for the reaction between propyne (C₃H₄) and bromine (Br₂) is:

C₃H₄ + 2Br₂ → C₃H₄Br₄


Molar mass of propyne (C₃H₄) = 3(12) + 4(1) = 36 + 4 = 40 g/mol

Number of moles of propyne = mass of propyne / molar mass of propyne
= 10 g / 40 g/mol
= 0.25 mol

From the balanced equation, we see that 1 mole of propyne requires 2 moles of bromine.

Number of moles of bromine required = 2 × number of moles of propyne
= 2 × 0.25 mol
= 0.5 mol

Molar mass of bromine (Br₂) = 2(80) = 160 g/mol

Mass of bromine required = number of moles of bromine × molar mass of bromine
= 0.5 mol × 160 g/mol
= 80 g

The correct answer is:

D. 80 g

vickyf

9 years ago

d ansa is correct....80g..... it is jst dat in d reaction d product was not properly written

Betty oluwatosin

11 years ago

Pls can u explain where d hydrogen came 4rm and d carbon attached to bromine

Azubuko1480

1 year ago

de answer d is not correct
some students might think it is correct those that got de correct answer will now choose de wrong answer due to de answer dere is d but de correct answer is b okay

boi12345

5 years ago

supposed to be cbr4 not c3br4
carbon tetra bromine

Stacy2009

1 year ago

I am confused

TheLivingLegend01

3 years ago

The balanced chemical equation for the reaction of propyne (C3H4) with bromine (Br2) is:

C3H4 + Br2 -> C3H4Br2

From the equation, we can see that 1 mole of propyne reacts with 1 mole of bromine to produce 1 mole of 1,2-dibromopropane. The molar mass of propyne is:

3(12.01 g/mol) + 4(1.01 g/mol) = 40.04 g/mol

So, 10 g of propyne is equal to:

10 g / 40.04 g/mol = 0.2498 mol

According to the balanced equation, 1 mole of propyne reacts with 1 mole of bromine. Therefore, 0.2498 mol of bromine is required to react with 0.2498 mol of propyne. The molar mass of bromine is 2(79.90 g/mol) = 159.80 g/mol. Thus, the mass of bromine required is:

0.2498 mol x 159.80 g/mol = 39.96 g

Therefore, the answer is approximately 40 g of bromine (option B).