15.99 x 2 g
16 grams or 32 grams, depending on what you're looking at.The mass of one mole of oxygen atoms is about 16 grams, but oxygen gas exists as a molecule (O2) so 1 mole of (O2) is 32 grams.
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
You have :N2 + O2 -----> 2NOmoles NO = ( 2 mols N2 reacted ) ( 2 mol NO / mol N2 reacted )moles NO = 4 moles NOmass NO = ( 4 mol NO ) ( 30 g NO / mol NO ) = 120 g NO
24g H2 x 1 mol H2 x 2 mol H2O x 18g H2O/ 2g H2 x 2 mol H20= 216 g H2O 160g O2 x 1 mol O2 x 2 mol H2O x 18g H2O/ 32g O2 x 1 mol O2 x 1 mol H2O= 180 g H2O
12.54 (g O2) / 2*15.99 (g/mol O2) = 0.3921 mol O2 -->0.3921 (mol O2) * 6.022*1023 (molecules O2)/(molO2) == 2.361*1023 molecules O2= 4.723*1023 atoms O
Ar of O = 16g/mol Mr of O2 = 2(16) = 32g/mol Using the formula : mass = Mr x number of moles mass = 32g/mol x 50mols = 1600g
Because oxygen gas (O2) has a molar mass of 32g/mol, 11.3 g * 1/32 mol/g gives about .35 moles. An ideal gas has a volume of 22.4 L/mol at STP, so 11.3 g O2 would have a volume of 7.91 L at STP.
The mass of 2.000 mol of oxygen atoms is 32.00 grams.
16 grams or 32 grams, depending on what you're looking at.The mass of one mole of oxygen atoms is about 16 grams, but oxygen gas exists as a molecule (O2) so 1 mole of (O2) is 32 grams.
Yes. CO2 has a weight of 44g/mol and O2 has a weight of 32g/mol.
228.37092g/mol
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
an equation showing conservation of mass of reactants and products:2H2O --> 2H2 & O2what is the mass of the oxygen gas produced, from 178.8 g H2O , (using molar masses:178.8 g H2O @ (1mol O2)(@ 32.00 g/mol) / (2molH2O)(18.02g/mol) =your answer (4 sigfigs): 158.8 g O2178.8 g H2O ---> 20.0 H2 & 158.8 O2
16.0 lbs = 7,257 g of O2 Oxygen exists in the gas form as 02 7,257g O2 * 1 mol O2/ 32 g * 2 mol O / 1 mol O2 * 6.022x10^23 atoms/mol = 2.73E26 atoms of O
The mass of O2 is approximately 32.00g/mol. The mass of H2 is about 2.016g/mol. These masses can be obtained by using the atomic weights of the atoms that make up these molecules.
55 grams O2 × (1 mol O2 ÷ 32 g O2) × (2 mol H2O ÷ 1 mol O2) × (22.4 L ÷ 1 mol H2O) = 77 L H2O(g)In the liquid form, it works out to just under 62 milliliters.
You have :N2 + O2 -----> 2NOmoles NO = ( 2 mols N2 reacted ) ( 2 mol NO / mol N2 reacted )moles NO = 4 moles NOmass NO = ( 4 mol NO ) ( 30 g NO / mol NO ) = 120 g NO