Consider the reaction: CH4 + 2 O2 → CO2 + 2 H2O
Select the true statement:
1. 4.00 moles of O2 can react to produce 4.00
moles of CO2.
2. 4.00 moles of CH4 will react with 2.00
moles of O2.
3. 8.00 moles of O2 can react to produce
144.0 g of H2O.
4. The limiting product is CO2.
5. 88.01 g of CO2 could be produced upon
reaction of 64.00 g of O2.
6. The limiting reactant is CO2

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally