Answer:
There is 52.33 grams of water produced.
Explanation:
Step 1: Data given
Mass of propane burned = 32.00 grams
Molar mass of propane = 44.1 g/mol
Oxygen is in excess
Molar mass of water = 18.02 g/mol
Step 2: The balanced equation
C3H8 + 5O2 → 4H2O + 3CO2
Step 3: Calculate moles of propane
Moles of propane = mass propane / molar mass of propane
Moles of propane = 32.00 grams / 44.1 g/mol
Moles of propane = 0.726 moles
Step 4: Calculate moles of H2O
Propane is the limiting reactant.
For 1 mol of propane consumed, we need 5 moles of O2 to produce 4 moles of H2O and 3 moles of CO2
For 0.726 moles of propane we'll have 4*0.726 = 2.904 moles of H2O
Step 5: Calculate mass of H2O
Mass of H2O = moles of H2O * molar mass of H2O
Mass of H2O = 2.904 moles * 18.02 g/mol
Mass of H2O = 52.33 grams
There is 52.33 grams of water produced.
When an object enters the Earth's atmosphere, it experiences a few forces, including gravity and drag. Gravity will naturally pull an object back to earth.
-Hope this helped, have a great day. ;)
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Answer:
12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.
Explanation:
For first solution of sulfuric acid :
C₁ = 40% , V₁ = ?
For second solution of sulfuric acid :
C₂ = 10% , V₂ = ?
For the resultant solution of sulfuric acid:
C₃ = 28% , V₃ = 20L
Also,
<u>V₁ + V₂ = V₃ = 20L</u> ......................................(1)
Using
<u>C₁V₁ + C₂V₂ = C₃V₃</u>
<u>40×V₁ + 10×V₂ = 28×20</u>
So,
40V₁ + 10V₂ = 560........................................(2)
Solving 1 and 2 as:
V₂ = 20 - V₁
Applying in 2
40V₁ + 10(20 - V₁) = 560
40V₁ + 200 - 10V₁ = 560
30V₁ = 360
<u>V₁ = 12 L</u>
So,
<u>V₂ = 20 - V₁ = 8L</u>
<u><em>12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.</em></u>
The molecular formula of methylpropan-1-ol is C4H10O, so the complete combustion equation is: C4H10O + 6O2 --> 4CO2 + 5H2O. This mean to completely combust 1.0mol of methylpropan-1-ol, 6 mol of O2 is required. Molar mass of O2 is 32 g/mol, so 32g/mol x 6mol = 192 g of O2 is required. At room temperature and pressure, the density of O2 is 1.3315 g/L (this can be obtained by density of gas = P/RT). So the volume of O2 = mass/density = 192g/1.3315(g/L) = 144 L = 144 dm3. The answer is B.
