Density is found by dividing mass over volume:
d=M/V. In this problem, we know the density, and the mass. Solve the general equation for volume, then enter the values from the problem and evaluate:
d=m/v [multiply v to both sides, then divide d from both sides]
v=m/d
v=83.8g/(2.33g/cm³)
v=35.965 cm³
v=36.0 cm³ to three significant figures (since your given information only has 3 sig figs)
Answer:
Percent yield = 50%
Explanation:
Given data:
Mass of CH₄ = 16 g
Mass of O₂ = 32 g
Mass of CO₂ = 11 g
Percent yield of CO₂ = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass/ molar mass
Number of moles = 16 g /16 g/mol
Number of moles = 1 mol
Number of moles of O₂:
Number of moles = mass/ molar mass
Number of moles = 32 g /32 g/mol
Number of moles = 1 mol
Now we will compare the moles of CO₂ with both reactant.
O₂ : CO₂
2 : 1
1 : 1/2×1= 0.5 mol
CH₄ : CO₂
1 : 1
Number of moles of CO₂ produced by oxygen are less so it will limiting reactant.
Theoretical yield:
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 0.5 mol × 44 g/mol
Mass = 22 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 11 g/ 22 g × 100
Percent yield = 50%
Answer:
Combustion is the reaction type if you meant to put a "=" in between C3H2 and H2O
