Qualitative analysis

Question

What volume of steam is produced when 10g of propyne is burnt in excess oxygen at stp?

Answer 1

Balanced reaction:

C₃H₄  + 4O₂  --> 3CO₂  + 2H₂O

 

Molar mass of propyne = 3*12.01g/mol Carbon + 4*1.01g/mol Hydrogen

= 40.07g/mol C₃H₄

 

10g C₃H₄ * 1 mol C₃H₄ / 40.07g = 0.25 mol C₃H₄

 

Stoichiometry:

2 mol H₂O is produced for every 1 mol C₃H₄

 

0.25 C₃H₄ *(2 mol H₂O / 1 mol C₃H₄) = 0.5 mol H₂O produced

 

Ideal Gas Law:

PV = nRT

 

(STP) Standard Temperature and Pressure  

(1 atm)*V = (0.5 mol H₂O)(0.08206 L atm mol ^-1 K^-1) (0.08206 L atm mol¯1 K¯1) (273.15 K)

V = 11.21 L H₂O

Answer 2

First you will have to write your balanced equation:

propyne+oxygen-> water+ carbon dioxide

 

C3H4 + 4 O2 -> 2 H2O + 3 CO2

 

You are looking to take the 10g of propyne and determine the amount of water produced

 

10g propyne * (1mol/ 40.06g propyne) * (2 mol water/ 1 mol propyne) * (22.4 L gas / 1 mol) = 11.18 L steam 

 

you will need to adjust for correct sig figs.

Let me know if you have further questions!