Are masses of $0.24 \cdot g$ $0.24g$ $M g$ $Mg$ metal and $0.64 \cdot g$ $0.64g$ $O_{2}$ $O_2$ gas stoichiometric with respect to the formation of $M g O$ $MgO$ ?

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Are masses of $0.24 \cdot g$ $0.24g$ $M g$ $Mg$ metal and $0.64 \cdot g$ $0.64g$ $O_{2}$ $O_2$ gas stoichiometric with respect to the formation of $M g O$ $MgO$ ?

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Answer 1 · 2017-10-05T05:37:43

$M g (s) + \frac{1}{2} O_{2} (g) \to M g O (s)$ $Mg(s) + 1/2O_2(g) rarrMgO(s)$

Explanation:

The stoichiometric reaction shows that $24.3 \cdot g$ $24.3*g$ reacts with $16.00 \cdot g$ $16.00*g$ of dioxygen gas to give approx. $40 \cdot g$ $40*g$ of the oxide.....

And for the quoted masses we can interrogate the molar quantities....

$Moles of magnesium = \frac{0.24 \cdot g}{24.3 \cdot g \cdot m o l^{- 1}} = 0.01 \cdot m o l$ $"Moles of magnesium"=(0.24*g)/(24.3*g*mol^-1)=0.01*mol$ .

$Moles of dioxygen = \frac{0.64 \cdot g}{32.0 \cdot g \cdot m o l^{- 1}} = 0.02 \cdot m o l$ $"Moles of dioxygen"=(0.64*g)/(32.0*g*mol^-1)=0.02*mol$ .

And thus a 2:1 molar ratio as required by the stoichiometry.

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Are masses of $0.24 \cdot g$ $0.24g$ $M g$ $Mg$ metal and $0.64 \cdot g$ $0.64g$ $O_{2}$ $O_2$ gas stoichiometric with respect to the formation of $M g O$ $MgO$ ?

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Explanation:

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Are masses of 0.24*g0.24⋅g0.24*g MgMgMg metal and 0.64*g0.64⋅g0.64*g O_2O2O_2 gas stoichiometric with respect to the formation of MgOMgOMgO?

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Explanation:

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Are masses of $0.24 \cdot g$ $0.24g$ $M g$ $Mg$ metal and $0.64 \cdot g$ $0.64g$ $O_{2}$ $O_2$ gas stoichiometric with respect to the formation of $M g O$ $MgO$ ?