Answer:
The mass of oxygen gas required to produce 65.75 grams of steam is approximately 162.2 grams
Explanation:
From the question, we have the following chemical reaction equation;
2C₃H₁₈(l) + 25O₂ (g) → 16CO₂(g) + 18H₂O (g)
The molar mass of oxygen, O₂ = 32 g/mol
The molar mass of steam, H₂O = 18.01528 g/mol
25 moles of oxygen are required to produce 18 moles of steam
Therefore, according to Proust's law of definite proportions;
(32 × 25) g of oxygen are required to produce (18 × 18.01528) g of steam
65.75 g of steam will be produced by (32 × 25)/(18 × 18.01528) × 65.75 g ≈ 162.2 g of oxygen O₂.
Answer : The number of moles present in ammonia is, 70.459 moles.
Solution : Given,
Mass of ammonia = 
Molar mass of ammonia = 17.031 g/mole
Formula used :
Therefore, the number of moles present in ammonia is, 70.459 moles.
For equal moles of gas, temperature can be calculated from ideal gas equation as follows:
P×V=n×R×T ...... (1)
Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.
2.4 atm ×3.25 L=n×R×297.5 K
Rearranging,
n\times R=0.0262 atm L/K
Similarly at final pressure and volume from equation (1),
1.5 atm ×4.25 L=n×R×T
Putting the value of n×R in above equation,
1.5 atm ×4.25 L=0.0262 (atm L/K)×T
Thus, T=243.32 K
