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How many moles are in 95.0 gram of octane?
how many moles are in 95.0 gram of octane?
How many moles of carbon and hydrogen are there in 5 moles of octane?
In 5 moles of octane, C8H18, there are 40 moles of carbon atoms (5 moles octane x 8 carbon atoms) and 90 moles of hydrogen atoms (5 moles octane x 18 hydrogen atoms).
How many moles of are present in 16.0 of octane?
To determine how many moles of octane are present in 16.0 g, you would divide the mass of octane by its molar mass. The molar mass of octane (C8H18) is approximately 114.23 g/mol. Therefore, 16.0 g ÷ 114.23 g/mol = 0.14 moles of octane.
How many moles of H2O are produced when 0.721mol of octane is burned?
6,49 moles of water are obtained.
How many moles of water will be produced from 1 mole of octane?
9
How many moles of carbon dioxide are formed in the complete combustion of one mole of octane C8H18?
The answer is 8 moles CO2.
7. How many moles of water can be produced by burning 325 g of octane (C8H18) in excess oxygen.?
To determine the moles of water produced from burning 325 g of octane (C8H18), we first need the balanced combustion reaction: [ \text{C}8\text{H}{18} + 12.5 \text{O}_2 \rightarrow 8 \text{CO}_2 + 9 \text{H}_2\text{O} ] Next, we calculate the molar mass of octane, which is approximately 114.22 g/mol. Thus, the number of moles of octane in 325 g is: [ \text{moles of C8H18} = \frac{325 \text{ g}}{114.22 \text{ g/mol}} \approx 2.85 \text{ moles} ] From the balanced equation, 1 mole of octane produces 9 moles of water. Therefore, the moles of water produced are: [ 2.85 \text{ moles C8H18} \times 9 \text{ moles H2O/mole C8H18} \approx 25.65 \text{ moles H2O} ] Thus, approximately 25.65 moles of water can be produced.
What is the Delta H value for liquid octane?
10,800 kJ (per 2 moles of octane; the balanced chemical equation)
How many moles of water can be produced by burning 325g of octane in excess oxygen?
Balanced equation: 2C8H18 + 25O2 ==> 16CO2 + 18H2Omoles of octane used: 325 g x 1 mole/114g = 2.85 moles octanemoles H2O produced: 18 moles H2O/2 moles C8H18 x 2.85 moles C8H18 = 25.65 moles H2O
How many moles of H2O are produced when 0.468 mol of octane is burned?
First write down the BALANCED reaction eq'n. Octane + oxygen = Water + Carbon Dioxide. 2CH3(CH2)6CH3 + 25O2 = 18H2O + 16CO2 The molar ratios are 2:25 :: 18:16 So '2' moles of octane produces 18 moles of water. So by equivlance 0.468 : x :: 2 : 18 Algebraically rearrgange x = (0.468/2) X 18 => x = 0.234 x 18 = > x = 4.212 moles water produces.
How much co2 is formed from the burning of 1.8 kg of octane?
The formula for normal octane is C8H10. Each mole burned creates 8 moles of CO2. A mole of octane is 106 grams, 8 moles of CO2 is 8x44 = 352 grams So 1.8 kg of octane would produce 1.8x(352/106) = 5.98 g CO2. As the octane value is given to 1 decimal place the answer can be no more accurate so 6.0 kg.
0.400mol of octane is allowed to react with 0.800mol of oxygen Which is the limiting reactant?
To determine the limiting reactant, compare the moles of each reactant to the stoichiometry of the balanced. In this case, the balanced equation is: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O The moles ratio between octane (C8H18) and oxygen (O2) is 2:25. Calculate the ratio for each reactant: Octane: 0.400 mol * (25 mol O2 / 2 mol C8H18) = 5.00 mol O2 needed Oxygen: 0.800 mol O2. Since the actual moles of oxygen available (0.800 mol) are greater than the moles needed for the reaction with octane (5.00 mol), oxygen is in excess and octane is the limiting reactant.
