3.61 *10^3 mol CO2
The answer is 8 moles CO2.
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.
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
Using the stoichiometry of the reaction, the molar ratio of C8H18 to CO2 is 2:16. Therefore, if 4.000 moles of C8H18 are consumed, 16/2 * 4.000 moles of CO2 will be produced. This results in 32.000 moles of CO2, which can be converted to grams using the molar mass of CO2 to get the theoretical yield in grams.
The balanced chemical equation for the combustion of carbon is: C + O2 → CO2 Calculate the moles of carbon and oxygen using their molar masses. Moles of carbon = 3.0g / 12.01 g/mol Moles of oxygen = 25.0g / 16.00 g/mol Since the reaction is 1:1 between carbon and oxygen, 1 mole of carbon reacts with 1 mole of oxygen to form 1 mole of carbon dioxide. Therefore, the mass of carbon dioxide formed would be the same as the mass of carbon burned, which is 3.0g.
The answer is 8 moles CO2.
2 moles C8H18 (18 moles H/1 mole C8H18) = 36 moles of hydrogen =================
The equation for a complete combustion reaction of CH4 is : CH4 + 2 O2 = CO2 + 2 H2O, showing that one mole of carbon dioxide is formed for each mole of CH4 burned. Therefore, the answer is 44 moles of CO2 formed.
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.
Out of one mole C3H8 three moles carbondioxide (CO2) are formed by complete combustion of it.So 2.13 mole C3H8 make 3*2.13 mole CO2 = 6.39 moleCO2
To determine the number of hydrogen atoms in 2 mol of C8H18, we first need to calculate the molar mass of C8H18. Carbon has a molar mass of approximately 12 g/mol, and hydrogen has a molar mass of approximately 1 g/mol. Therefore, the molar mass of C8H18 is (812) + (181) = 114 g/mol. Next, we use Avogadro's number (6.022 x 10^23) to calculate the number of molecules in 2 mol of C8H18, which is 2 mol * 6.022 x 10^23 molecules/mol. Since there are 18 hydrogen atoms in each molecule of C8H18, the total number of hydrogen atoms in 2 mol of C8H18 is 2 mol * 6.022 x 10^23 molecules/mol * 18 atoms/molecule = 2.17 x 10^25 hydrogen atoms.
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.
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
There are two words that can be formed using the letters in "wemoll": "mellow" and "mole".
Using the stoichiometry of the reaction, the molar ratio of C8H18 to CO2 is 2:16. Therefore, if 4.000 moles of C8H18 are consumed, 16/2 * 4.000 moles of CO2 will be produced. This results in 32.000 moles of CO2, which can be converted to grams using the molar mass of CO2 to get the theoretical yield in grams.
The balanced chemical equation for the combustion of carbon is: C + O2 → CO2 Calculate the moles of carbon and oxygen using their molar masses. Moles of carbon = 3.0g / 12.01 g/mol Moles of oxygen = 25.0g / 16.00 g/mol Since the reaction is 1:1 between carbon and oxygen, 1 mole of carbon reacts with 1 mole of oxygen to form 1 mole of carbon dioxide. Therefore, the mass of carbon dioxide formed would be the same as the mass of carbon burned, which is 3.0g.
Is there more to this question? You're asking for numbers for two different ideas without sufficient information. A mole [mol] is the molecular wgt in grams of a given substance which you haven't specified unless I misunderstand your question. Learn to ask questions with scrupulous accuracy. Often then, the answer will crystallize for you from that alone---ie; a properly framed question.