Answer:
0.800 mol
Explanation:
We have the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with moles of the compounds involved.
Step 1. <em>Gather all the information</em> in one place.
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 4.00 4.00
===============
Step 2. Identify the <em>limiting reactant
</em>
Calculate the <em>moles of CO₂</em> we can obtain from each reactant.
<em>From C₃H₈:</em>
The molar ratio of CO₂: C₃H₈ is 3:1
Moles of CO₂ = 4.00 × 3/1
Moles of CO₂ = 12.0 mol CO₂
<em>From O₂</em>:
The molar ratio of CO₂: O₂ is 3:5.
Moles of CO₂ = 4.00 × ⅗
Moles of CO₂ = 2.40 mol CO₂
O₂ is the limiting reactant because it gives the smaller amount of CO₂.
==============
Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.
The molar ratio of C₃H₈:O₂ is 1:5.
Moles of C₃H₈ = 4.00 × ⅕
Moles of C₃H₈ = 0.800 mol C₃H₈
Answer:
about 10 pounds
Explanation:
your mass doesn't change but weight changes (due to gravity)
To solve for the absolute temperature, we assume ideal gas
behaviour so that we use the equation:
PV = nRT
or T = PV / nR
So calculating:
T = [6.6 atm * 0.40 L] / [(2.4g / 28g/mol) * 0.08205746 L
atm / mol K]
<span>T = 375.35 K</span>
