How many grams of O2 are consumed when 10.0 grams of C12H22O11 or sugar are burned?

We know the amount of O2 consumed in this reaction in units of moles and we can calculate the mass of 0.3505 moles of O2 from the molecular weight of oxygen. According to this calculation, it takes 11.2 grams of O2 to burn 10.0 grams of sugar.

Which is the limiting reactant when 100.00 g sucrose C12H22O11 and 100.0 g oxygen O2 react to form water and carbon dioxide?

Oxygen is the limiting reagent.

What is δe for the combustion of 1 mole of sucrose C12H22O11?

Problem Details. The combustion of sucrose (a carbohydrate) is shown as follows C12H22O11(s) + 12O2(g) → 12 CO2(g) + 11H2O(ℓ) Calculate the work, ΔH°at 298 K, and ΔE° for the combustion of one mole of sucrose at 25°C and pressure. ΔHfsucrose = -2218 kJ/mol.

What would be the products of sucrose C12H22O11 being heated in the presence of oxygen?

3) When sucrose (C12H22O11) burns in oxygen, carbon dioxide, water and heat are produced.

What is the combustion of sucrose?

Combustion (metabolism) of sucrose (C12H22O11) to produce CO2 and H2O. 1 mole of C12H22O11 molecules produces 12 moles of CO2 molecules and 11 moles of H2O molecules. 1 mole of C12H22O11 molecules combines with 12 moles of O2 molecules.

How many grams of oxygen are there in 100 g of sucrose?

51.41 g
This means that 100.00 g of sucrose always contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen.

How do you find limiting reactant?

The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation.

How do you find the heat of combustion of sucrose?

The heat released upon combustion of the sample of sucrose can be calculated from the heat capacity c and the temperature change ∆ T ∆T ∆T : Q = c ∆ T = 3.60 ⋅ ( 28.28 − 21.40 ) = 24.768 Q = c∆T = 3.60·(28.28-21.40) = 24.768 Q=c∆T=3.

What is the heat of combustion of sucrose in kJ mol?

is -5645 kJ/mol.

What is the combustion reaction of sucrose?

Combustion of sucrose is given by the chemical equation, C12H22O11 + 12 O2 ——> 12CO2 + 11 H2O…

How do you find the combustion of sucrose?

Question: For reaction 2, the combustion of sucrose, the balanced equation is: C12H22O11 (s) + 12 O2 (g) → 12 CO2 (g) + 11 H2O (l) The mass used, and the temperature change are given in the experimental section.

How much oxygen is produced when sucrose reacts with C12H22O11?

For the combustion of sucrose: C12H22O11 + 12O2 → 12CO2 + 11H2O There are 12.0 g of sucrose and 10.0 g of oxygen reacting in lab conditions. How much H2O will be produced? (You will need notebook paper to solve this)

Does sucrose run out of oxygen before or after sucrose?

Oxygen on hand ⇒ 10.0 g / 31.9988 g/mol = 0.3125 mol Since the oxygen required is greater than that on hand, it will run out before the sucrose. Oxygen is the limiting reagent. Solution path #2: 1) Calculate moles:

How many moles of oxygen are required for sucrose to dissolve?

From the coefficients, we see that 12 moles of oxygen are require for every one mole of sucrose. Therefore: 0.0292146 mol times 12 = 0.3505752 mole of oxygen required 3) Determine limiting reagent: Oxygen on hand ⇒ 10.0 g / 31.9988 g/mol = 0.3125 mol

What is the balanced formula and reaction type for sucrose?

Sucrose, C12H22O11, is burned to produce carbon dioxide and water. What is the balanced formula and reaction type? How many carbon atoms are contained within 10.0 pounds of sugar, (sucrose) which has the formula c12h22o11? The molar mass of sucrose, or table sugar, is 342 grams.