CH4 + 2 H2O = 3 H2 + CO2
8 moles CH4 produce 8 x 3 moles H2, which is 24.
There are 4 Hydrogen atoms in 1 mol of CH4 so in 2 moles there will be 8.
2 moles CH4 (6.o22 X 10^23/1 mole CH4)
= 1.0 X 10^24 moles of methane
One mole has avagadro number(6.022x10^23).So two moles have twice
Take the balanced equation CH4+2O2---->CO2+2H2O. 16g of CH4 gives 36g of water.So 244g gives 549g of water
6.02 x 1024 atoms
1.2044x1024 molecules
1.5
10
NH3 Molecules = ( 8.1 x 10^20 H atoms ) ( 1 NH3 molecule / 3 H atoms ) NH3 Molecules = 2.7 x 10^20 NH3 molecules NH3 moles = ( NH3 molecules ) / ( N Avogadro ) NH3 moles = ( 2.7 x 10^20 NH3 molecules ) / ( 6.022 x 10^23 molecules / mole ) NH3 moles = 4.48 x 10^-4 NH3 moles <--------------
20 moles of NaOH needed to neutralize 20 moles of nitric acid
One "mole" of any substance is "Avogadro's number", of these atoms or molecules. Avogadro's Number is 6.023*10e23, or 6023 followed by another 20 zeroes. So 6.10 moles of sulfur is 6.1 * 6.023*10e23.
Find out the percentage of hydrogen in the molar mass of methane. Molar mass of CH4: C = 1 * 12.01 g = 12.01 g H = 4 * 1.01 g = 4.04 g Total = 16.05 g 4.04 g/16.05 g * 100% = 25.171% 0.25171 * 20 g = 5.0342 g There are about 5.03 grams of hydrogen in 20 grams of methane gas.
Molarity = moles of solute/Liters of solution Or, for our purposes, Moles of solute = Liters of solution * Molarity Moles Na2CO3 = 10.0 Liters * 2.0 M = 20 moles Na2CO3 --------------------------
To find moles, simply divide the number of representative particles (in this case, molecules of methane) by Avogadro's number (6.02x1023.)2.45x1023/6.02x1023 = approx. 0.41 moles (the exponents cancel out.)
NH3 Molecules = ( 8.1 x 10^20 H atoms ) ( 1 NH3 molecule / 3 H atoms ) NH3 Molecules = 2.7 x 10^20 NH3 molecules NH3 moles = ( NH3 molecules ) / ( N Avogadro ) NH3 moles = ( 2.7 x 10^20 NH3 molecules ) / ( 6.022 x 10^23 molecules / mole ) NH3 moles = 4.48 x 10^-4 NH3 moles <--------------
20 moles
0.2 moles C6H12O6 x 6.02x10^23 molecules/mole = 1.2x10^23 molecules of C6H12)61.2x10^23 molecules C6H12O6 x 6 molecules "O"/molecule C6H12O6 = 7.2x19^23 molecules "O"
20 moles of NaOH needed to neutralize 20 moles of nitric acid
There are 0.13 moles in 20 grams of magnesium nitrate.
One "mole" of any substance is "Avogadro's number", of these atoms or molecules. Avogadro's Number is 6.023*10e23, or 6023 followed by another 20 zeroes. So 6.10 moles of sulfur is 6.1 * 6.023*10e23.
Find out the percentage of hydrogen in the molar mass of methane. Molar mass of CH4: C = 1 * 12.01 g = 12.01 g H = 4 * 1.01 g = 4.04 g Total = 16.05 g 4.04 g/16.05 g * 100% = 25.171% 0.25171 * 20 g = 5.0342 g There are about 5.03 grams of hydrogen in 20 grams of methane gas.
No, this would make 5 moles. This is because water is H2O. This means that for each oxygen molecule used, there will be 2 hydrogen molecules used. In the given equation Only 5 moles of oxygen could be used to pair with all 10 moles of hydrogen, therefore giving you an excess of 5 oxygen molecules.
40 moles of LiOH
The molar mass of an element is its atomic weight in grams. The atomic weight is on the periodic table. 1 mole C = 12.0107 g C. To calculate the number of moles in 80 g of C, do the following: 80 g C x (1 mole C/12.0107 g C) = 6.66 mole C = 7 mole C* *This answer has been rounded to the proper number of significant figures. When multiplying or dividing, the answer is rounded to the fewest significant figures used in the calculation. 80 only has one significant figure. Refer to the related link below concerning significant figures.
Molarity = moles of solute/Liters of solution Or, for our purposes, Moles of solute = Liters of solution * Molarity Moles Na2CO3 = 10.0 Liters * 2.0 M = 20 moles Na2CO3 --------------------------