A $5 \cdot g$ $5g$ mass of magnesium metal reacts with a $4.0 \cdot g$ $4.0g$ mass of oxygen gas. Which reagent is in stoichiometric excess?

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A $5 \cdot g$ $5g$ mass of magnesium metal reacts with a $4.0 \cdot g$ $4.0g$ mass of oxygen gas. Which reagent is in stoichiometric excess?

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Answer 1 · 2017-02-21T08:40:15

We follow the stoichiometric equation:

$2 M g (s) + O_{2} (g) \to 2 M g O (s)$ $2Mg(s) + O_2(g) rarr 2MgO(s)$

Explanation:

$Moles of metal = \frac{5.0 \cdot g}{24.3 \cdot g \cdot m o l^{- 1}} = 0.206 \cdot m o l$ $"Moles of metal"=(5.0*g)/(24.3*g*mol^-1)=0.206*mol$

$Moles of dioxygen = \frac{4.0 \cdot g}{32.0 \cdot g \cdot m o l^{- 1}} = 0.125 \cdot m o l$ $"Moles of dioxygen"=(4.0*g)/(32.0*g*mol^-1)=0.125*mol$

$Dioxygen gas$ $"Dioxygen gas"$ is in stoichiometric excess, and only $0.103$ $0.103$ mol of dioxygen will react. And thus $0.022 \cdot m o l$ $0.022*mol$ dioxygen will remain unreacted, i.e. approx. $0.7 \cdot g$ $0.7*g$ .

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A $5 \cdot g$ $5g$ mass of magnesium metal reacts with a $4.0 \cdot g$ $4.0g$ mass of oxygen gas. Which reagent is in stoichiometric excess?

1 Answer

Explanation:

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A 5⋅g5*g mass of magnesium metal reacts with a 4.0⋅g4.0*g mass of oxygen gas. Which reagent is in stoichiometric excess?

1 Answer

Explanation:

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A $5 \cdot g$ $5g$ mass of magnesium metal reacts with a $4.0 \cdot g$ $4.0g$ mass of oxygen gas. Which reagent is in stoichiometric excess?