Y’all Someone plz help me with these problems.

Girl I will report if u troll or link

A. 650 moles of sulphur.

B. 30 g of FeS₂

Explanation:

The balanced equation for the reaction is given below:

Fe + 2S —> FeS₂

From the balanced equation above,

1 mole of Fe reacted with 2 moles of S to produce 1 mole of FeS₂.

A. Determination of the mole of sulphur needed for the reaction.

From the balanced equation above,

2 moles of S reacted to produce 1 mole of FeS₂.

Therefore, Xmol of S will react to produce 325 moles of FeS₂ i.e

Xmol of S = 2 × 325

Xmol of S = 650 moles

Thus, 650 moles of sulphur are needed for the reaction.

B. Determination of the mass of FeS₂ produced by the reaction of 0.5 mole of sulphur.

From the balanced equation above,

2 moles of S reacted to produce 1 mole of FeS₂.

Therefore, 0.5 mole of S will react to produce = (0.5 × 1)/2 = 0.25 mole of FeS₂.

Finally, we shall determine the mass of 0.25 mole of FeS₂. This can be obtained as follow:

Mole of FeS₂ = 0.25 mole

Molar mass of FeS₂ = 56 + (32×2)

= 56 + 64

= 120 g/mol

Mass of FeS₂ =?

Mass = mole × molar mass

Mass of FeS₂ = 0.25 × 120

Mass of FeS₂ = 30 g

Thus, 30 g of FeS₂ were obtained from the reaction.

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