1.89 g H2 is allowed to react with 9.91 g N2, producing 2.37 g NH3. Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.

The theoretical mass of NH3 is 10.6 grams

Explanation:

Step 1: Data given

Mass of H2 = 1.89 grams

Mass of N2 = 9.91 grams

Mass of NH3 = 2.37 grams = actual yield

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

-Step 3: Calculate moles

Moles = mass / molar mass

Moles H2 = 1.89 grams / 2.02 g/mol

Moles H2 = 0.936 moles

Moles N2 = 9.91 grams / 28.0 g/mol

Moles N2 = 0.354 moles

Step 4: Calculate the limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

H2 is the limiting reactant. It will completely be consumed, 0.936 moles

N2 is in excess. There will react 0.936/ 3 = 0.312 moles N2

There will remain 0.354 - 0.312 = 0.042 moles

Step 5: Calculate moles NH3

For 3 moles H2 we'll have 2 moles NH3

For 0.936 moles H2 we'll have 2/3 * 0.936 =  0.624 moles NH3

Step 6: Calculate mass NH3

Mass NH3 = 0.624 moles * 17.03 g/mol

Mass NH3 = 10.6 grams

The theoretical mass of NH3 is 10.6 grams

the answer is option a.

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