The balanced equation for the above reaction is as follows;
2Ca + O₂ ---> 2CaO
stoichiometry of Ca to O₂ is 2:1
we first need to find the limiting reactant
number of Ca moles - 6.84 mol
number of O₂ moles - 4.00 mol
if Ca is the limiting reactant
if 2 mol of Ca reacts with 1 mol of O₂
then 6.84 mol of Ca reacts with - 6.84 / 2 = 3.42 mol of O₂
this means that Ca is the limiting reactant and O₂ is in excess
therefore amount of CaO produced depends on amount of limiting reactant present
stoichiometry of Ca to CaO is 2:2
number of moles of Ca reacted = number of CaO moles formed
number of moles of CaO formed - 6.84 mol
answer is 6.84 mol
0.035 mol CF₄ x 4 mols F/ 1 mol CF₄ = 0.14 mols F
0.14 mols F x 6.022x10²³/ 1 mol F= 8.43x10²² atoms
answer: 8.4×10²² atoms of F
maybe try this im not 100% sure though
Answer: 10.7 moles
Explanation: Hydrogen gas exists as a diatomic gas (H2) Hence, the Molar mass of one mole of hydrogen gas is 1 x 2 = 2 g. To find the number of moles from the mass take the weight and divide is by the molar mass of the molecule. So in this case, 21.4/2 = 10.7 moles