19 What is the number of moles of nitrogen gas which react with 18 g of magnesium to form magnesium nitride compound ? [Mg = 24,N=14]
a) 0.25 mol 0 b) 0.5 mol C) 1 mol d) 2 mol​

Respuesta :

Answer:

0.5 mol of nitrogen gas.

Explanation:

18g of Magnesium in mol is equal to 18g/24 gmol-1 = 0.75 mol and 7g of nitrogen in mol is equal to 7g/14 gmol-1 = 0.5 mol,

Thus, When 18g of magnesium reacts with 7g of nitrogen it implies 0.75 mol of magnesium which reacts with 0.5 mol of nitrogen. :)

The number of moles of nitrogen gas which react with 18 g of magnesium

to form magnesium nitride compound is 0.5mol.

We need to find the number of mole for magnesium and nitrogen.

Mole= mass/molar mass

Magnesium= 18g/24gmol⁻¹ = 0.75mol of Mg

Nitrogen= 7g/14gmol⁻¹ = 0.5mol of N

This implies that when 18g of magnesium reacts with 7g of nitrogen it

means 0.75 mol of magnesium which reacts with 0.5 mol of nitrogen.

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