How much magnesium metal is needed to prepare a #25*g# mass of magnesium oxide?

Answer 1

Approx. #15*g# of magnesium metal.........

We need (i) a stoichiometric equation:

#Mg(s) + O_2(g) rarr MgO(s)#,

and (ii), equivalent quantities of the product #MgO#:

#"Moles of magnesium oxide"=(25*g)/(40.30*g*mol^-1)=0.620*mol#.

Now from the stoichiometry of the equation, if there are #0.620*mol# #MgO#, then there MUST have been at least #0.620*mol# #"magnesium metal"#, and this quantity represents a MASS of #0.620*molxx24.31*g*mol^-1=??g#

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Answer 2

Caleb Adams

The molar mass of magnesium oxide (MgO) is 40.3 g/mol. Since magnesium oxide has a 1:1 ratio of magnesium to oxygen atoms, the molar mass of magnesium (Mg) is 24.3 g/mol. To prepare 25 g of magnesium oxide, you need (25 g / 40.3 g/mol) * 24.3 g/mol = approximately 15.09 g of magnesium metal.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.