The Atomic Mass of the organic compound CH3OH is 12.0 + 3.0 + 16.0 + 1.0 = 32.0Amount of CH3OH = mass of pure sample/molar mass = 5.00/32.0 = 0.156molThere are 0.156 moles of CH3OH in a 5.00 gram pure sample.
200 g CH4 x 1 mole CH4/16 g = 12.5 moles CH4
Divide 96 by molecular mass.So the answer is 6mol
I. False - Since both gases contribute to the total mass, the number of moles of NO does not necessarily need to be greater than the number of moles of CH4. II. True - If the total mixture mass is 17 grams and CH4 is 8 grams, then the remaining mass must be of NO. III. True - If the total moles of the mixture is 0.8, and CH4 is 0.5 moles (8g/16 g/mol), then the moles of NO would be the remaining 0.3 moles.
To calculate the mass of oxygen required to react with 20 grams of CH4, we first need to write and balance the chemical equation for the reaction. The balanced equation for the combustion of CH4 is: CH4 + 2O2 → CO2 + 2H2O This equation tells us that 1 mole of CH4 reacts with 2 moles of O2. The molar mass of CH4 is 16 g/mol. Therefore, 20 grams of CH4 is equal to 20/16 = 1.25 moles CH4. So, 1.25 moles of CH4 would require 2.50 moles of O2. The molar mass of O2 is 32 g/mol. Therefore, the mass of O2 required would be 2.50 moles * 32 g/mol = 80 grams.
2,8 moles is of course equivalent to 2,8 moles !Probable is a spelling error in your question.
To convert moles of CH4 (methane) to grams, you would use the molar mass of CH4, which is approximately 16.04 g/mol. Multiply the number of moles of CH4 by this molar mass to obtain the mass in grams. The formula is: grams of CH4 = moles of CH4 × 16.04 g/mol.
200 g CH4 x 1 mole CH4/16 g = 12.5 moles CH4
Divide 96 by molecular mass.So the answer is 6mol
I. False - Since both gases contribute to the total mass, the number of moles of NO does not necessarily need to be greater than the number of moles of CH4. II. True - If the total mixture mass is 17 grams and CH4 is 8 grams, then the remaining mass must be of NO. III. True - If the total moles of the mixture is 0.8, and CH4 is 0.5 moles (8g/16 g/mol), then the moles of NO would be the remaining 0.3 moles.
To calculate the mass of oxygen required to react with 20 grams of CH4, we first need to write and balance the chemical equation for the reaction. The balanced equation for the combustion of CH4 is: CH4 + 2O2 → CO2 + 2H2O This equation tells us that 1 mole of CH4 reacts with 2 moles of O2. The molar mass of CH4 is 16 g/mol. Therefore, 20 grams of CH4 is equal to 20/16 = 1.25 moles CH4. So, 1.25 moles of CH4 would require 2.50 moles of O2. The molar mass of O2 is 32 g/mol. Therefore, the mass of O2 required would be 2.50 moles * 32 g/mol = 80 grams.
There are 0.75 moles in it.You have to devide 12 by molecular mass
For every 1 mole of CH4 that reacts, 1 mole of CO2 is produced. Therefore, 4 moles of CH4 will produce 4 moles of CO2. To calculate the mass of CO2 produced, you would need to multiply the moles of CO2 by its molar mass (44 g/mole) to get the total mass produced.
This mass is 6,416 g.
2,8 moles is of course equivalent to 2,8 moles !Probable is a spelling error in your question.
First, determine the heat of combustion for CH4, which is -802 kJ/mol. Then, calculate the number of moles of CH4 needed to emit 267 kJ of heat. Finally, convert the moles of CH4 to grams using the molar mass of CH4 (16 g/mol).
First, determine molar mass of CH4: C:12g/mol + 4x H:1g/mol= 16g/mol Then divide by the number of grams. 64g/(16g/mol)= 4 moles of CH4
Molar mass:O2 32 g/molCH4 16 g/molReaction:CH4 + 2 O2 --> CO2 + 2H2OCalculus1 mol CH4 with 2 moles O2Given: 24 g CH4 equals 24/16 = 1.5 mol CH4this will need 2 * 1.5 moles O2 = 3.0 mol O23.0 mol O2 equals 3.0 * 32 = 96 g O2