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
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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
The answer is C : 15.7 m/s
Use the idea of : momentum before collision = momentum after collision
Before collision;
For car:mass=1.1×10^3, velocity=22
For truck:mass=2.3×10^3, velocity=0
After collision;
For car:mass=2.3×10^3, velocity=-11
For truck:mass=2.3×10^3, velocity=V
(1.1×10^3 × 22) + (2.3×10^3 × 0) = (1.1×10^3 × -11) + (2.3×10^3 × V)
24200 = -12100 + 2.3×10^3V
2.3×10^3V = 36300
V = 15.7 m/s
Answer:
4Cr + 3O2 —> 2Cr2O3
Explanation:
Information from the question include:
Chromium + oxygen -> chromium(III) oxide
From the word equation given above, the equation can be written as follow:
Cr + O2 —> Cr2O3
The equation can be balance by doing the following:
There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of Cr2O3 and 3 in front of O2 as shown below:
Cr + 3O2 —> 2Cr2O3
Now, we have 4 atoms of Cr on the right side and 1 atom on the left. It can be balance by putting 4 in front of Cr as shown below:
4Cr + 3O2 —> 2Cr2O3
Now the equation is balanced
Answer:
3%
Explanation:
Substract the actual error from the final and multiply by 100
