B. Are produced only by living things
Nitrate 2 , Hydrogen Oxygen , and Carbon Oxide
Professor of Physics Carlo Rubbia
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
From the coefficients of the equation, we know that for every 3 moles of water consumed, 1 mole of diphosphorus trioxide is consumed.
This means we need to find the mass of 0.75 moles of diphosphorus trioxide.
- The atomic mass of phosphorous is 30.973761998 g/mol.
- The atomic mass of oxygen is 15.9994 g/mol.
So, the formula mass of diphosphorus trioxide is:
- 2(30.973761998)+3(15.9994)=109.945723996 g/mol.
Thus, 0.75 moles have a mass of:
- 0.75(109.945723996), which is about 82.5 g (to 3 sf)
