Question

(b) Calculate the number of moles in 2400 g of () CO, (ii) Oxygen molecule (iii) Caco, (iv) MgBr, ​

Answer 1

Answer:

oxygen module

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Answer 2

Answer:

(i) 85 Moles (ii) 75 Moles (iii) 35 Moles (iv) 23 Moles

Explanation:

`(i) Carbon Monoxide (CO)

Molecular mass of CO = 12 + 16 = 28 u

Molar mass of CO = 28 g

Number of moles = $\frac{2400}{28}$

= 85 Moles (Approx, real value is 85.7142857143 )

(ii) Oxygen ($O_{2}$)

Molecular mass of $O_{2}$ = 16 + 16 = 32 u

Molar mass of $O_{2}$ = 32 g

Number of moles = $\frac{2400}{32}$

= 75 moles

(iii) Calcium Carbonate (CaCO)

Molecular mass of CaCO = 40 + 12 + 16 = 68 u

Molar mass of CaCO = 68 g

Number of moles = $\frac{2400}{68}$

= 35 moles (Approx, real value is 35.2941176471 )

(iv) Magnesium Bromine (MgBr)

Molecular mass of MgBr = 24 + 80 = 104 u

Molar mass of MgBr = 104 g

Number of moles = $\frac{2400}{104}$

= 23 moles (Approx, real value is 23.0769230769 )

I hope this helps!!

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Lilac584