How many grams of HBr would exactly be required to react with 2 g of propyne? (C =12, H=1, Br = 80)
4. 1 g
6. 1 g
8. 1g
16.2g
Explanation
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2C3H4 + 4HBr ---> 2C3H2Br2
Frm Above Eqn
Mole Ratio Of C3H4 And HBr
Is 2:4 => 1:2
Then Number Of Mole=reacting mass/Molar Mass
Hint: C3H4=40g/mol
HBr=81g/mol
For Propyne
N=2/40 = 0.05mol
If The Ratio Of The Two Reactant Is 1:2
Then C3H4 1: HBr 2
0.05 : 0.10
Then Reacting Mass Of HBr ==> m= N x M
0.10x81
=8.1gms Of HBr
Here's how to solve this problem:
**1. Understand the Reaction:**
* Propyne (C₃H₄) is an alkyne, meaning it has a triple bond. HBr can add across a triple bond, potentially adding twice.
* Since the problem says "exactly required", we assume **complete addition**, meaning both possible HBr additions happen. This converts the triple bond into a single bond.
**2. Write the Balanced Chemical Equation:**
C₃H₄ + 2HBr → C₃H₆Br₂ (Propyne + 2 Hydrogen Bromide --> Dibromopropane)
**3. Calculate the Molar Mass of Propyne (C₃H₄):**
* (3 x 12) + (4 x 1) = 36 + 4 = 40 g/mol
**4. Calculate the Molar Mass of HBr:**
* 1 + 80 = 81 g/mol
**5. Calculate the Moles of Propyne:**
* Moles of propyne = Mass / Molar Mass = 2 g / 40 g/mol = 0.05 mol
**6. Determine the Moles of HBr Required:**
* From the balanced equation, 1 mole of propyne reacts with 2 moles of HBr.
* Therefore, 0.05 moles of propyne will react with 0.05 mol * 2 = 0.1 moles of HBr.
**7. Calculate the Mass of HBr Required:**
* Mass of HBr = Moles x Molar Mass = 0.1 mol x 81 g/mol = 8.1 g
**Therefore, the answer is C. 8.1 g**
My school is right :it is really 164
get the equation balanced, 2 is needed at the back of HBr
if the mass no of HBr is 81(1+80)*2 then is the no of moles of the compound that will react with propyne not meant to be162 and not 164? pls recheck the solving.except if the compound is H2Br. so the answer in this question based on what we are given is not in the options.
@MySchool.NG Just like Mandy Blessing Christabel penned, your answer should read 162 for clarity sake.
