Given the following: 2 C8H18 + 25 O2 ====è 16 CO2 + 18 H2O Find the...

Question

Question

Given the following:

2 C₈H₁₈ + 25 O₂ ====è 16 CO₂ + 18 H₂O

Find the limiting reagent for making CO₂ from 35.4 Moles C₈H₁₈ and 145 moles O2. Then find the grams excess reagent left at the end of the reaction

Answer 1

Answer #1

If you have any query please comment

If you satisfied with the solution please like it thankxx

Answer 2

Similar Homework Help Questions

Burning octane is: 2 (C8H18) + 25 (O2) ......... 16 (CO2) + 18 (H2O) How many...

Burning octane is: 2 (C8H18) + 25 (O2) ......... 16 (CO2) + 18 (H2O) How many moles of oxygen would you need to burn 2.37 moles of octane?
How many liters of CO2 would be produced from the reaction of 5.0 L of C8H18 with 55 L of O2?

2 C8H18 (l) + 25 O2 (g) ----> 16 CO2 (g) + 18 H2O (g)(OR 2 C8H18 (l) + 25 O2 (g) right arrow 16 CO2 (g) + 18 H2O (g))Use the following conversions: 1 mol = MW in g (2 d.p.) = 22.414 LAll gases are at STP. How many liters of CO2 would be produced from the reaction of 5.0 L of C8H18 with 55 L of O2?Theoretical yield of CO2 from 5.0 L C8H18 = _____________Theoretical yield...
Combustion of Octane: C8H18 + O2 → CO2 + H2O

Combustion of Octane: C8H18 + O2 → CO2 + H2O Question 1: What are the coefficients for the balanced reaction of the combustion of octane? Problem 2: If 20 g of octane combust with 20 g of oxygen, which is the limiting reagent?

The combustion of octane (C8H18) in oxygen proceeds as follows 2 C8H18(g) + 25 O2(g) --->...

The combustion of octane (C8H18) in oxygen proceeds as follows 2 C8H18(g) + 25 O2(g) ---> 16 CO2 (g) + 18 H2O(l) How many moles of CO2are produced when 5.0 moles of octane, C8H18, is burned in 5.0 moles of oxygen?
Given the following reaction: 4 Al + 3 O2 =====è 2 Al2O3 Find the grams O2...

Given the following reaction: 4 Al + 3 O2 =====è 2 Al2O3 Find the grams O2 used and the grams Al2O3 made with 855 grams Al and excess O2 Given the following reaction: 4 Al + 3 O2 ====> 2 Al2O3 Find the grams O2 used and the grams Al2O3 made with 855 grams Al and excess O2
Give the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18...

Give the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18 with 8.000 moles of O2.2 C8H18 + 25 O2 ? 16 CO2 + 18 H2Oa)12.50%b)25.00%c)20.00%d)50.00%

Part A Determine the balanced chemical equation for this reaction. C8H18(g)+O2(g)?CO2(g)+H2O(g) Enter the coefficients for each...

Part A Determine the balanced chemical equation for this reaction. C8H18(g)+O2(g)?CO2(g)+H2O(g) Enter the coefficients for each compound in order, separated by commas. For example, 1,2,3,4 would indicate one mole of C8H18, two moles of O2, three moles of CO2, and four moles of H2O. 2,25,16,18 SubmitHintsMy AnswersGive UpReview Part Correct It is important to balance a chemical equation before using it for calculations. Checking that equations are balanced will help you avoid many errors in chemistry problems. Balanced chemical equation...
The combustion of octane, C8H18, proceeds according to the reaction 2 C8H18(l) + 25 O2(g) -->...

The combustion of octane, C8H18, proceeds according to the reaction 2 C8H18(l) + 25 O2(g) --> 16 CO2(g) + 18 H2O(l) If 442 mol of octane combusts, what volume of carbon dioxide is produced at 20.0 ?C and 0.995 atm?
Octane (C8H18) reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O).

Octane (C8H18) reacts with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). When the equation in balanced, the coefficient of octane is: C8H18+ O2 --> CO2 + H2O a.8 b. 16 c.25 d. 2

the complete combustion of octane , C8H18, a component og gasoline, proceed as 2 C8H18 +25...

the complete combustion of octane , C8H18, a component og gasoline, proceed as 2 C8H18 +25 O2 = 16 CO2 + 18H2O. a) how many moles of CO2 are produced when 1.50 mol octance reacted? b) how many grams of water produced in this reaction? c)how many moles of oxygen required to form 90.0 g water?