consider this reaction.\n2mg(s)+o₂(g)→2mgo(s)\nwhat volume (in milliliters) of oxygen gas is required to…

Explanation:

Step1: Calculate moles of Mg

The molar - mass of Mg is $M_{Mg}=24.31\ g/mol$. The number of moles of Mg, $n_{Mg}$, is calculated using the formula $n=\frac{m}{M}$. So, $n_{Mg}=\frac{4.03\ g}{24.31\ g/mol}\approx0.166\ mol$.

Step2: Determine moles of $O_2$ from the stoichiometry

From the balanced chemical equation $2Mg(s)+O_2(g)\rightarrow2MgO(s)$, the mole - ratio of $Mg$ to $O_2$ is $n_{Mg}:n_{O_2}=2:1$. So, $n_{O_2}=\frac{1}{2}n_{Mg}$. Substituting $n_{Mg}=0.166\ mol$, we get $n_{O_2}=\frac{0.166\ mol}{2}=0.083\ mol$.

Step3: Calculate volume of $O_2$ at STP

At STP (Standard Temperature and Pressure, $T = 273\ K$ and $P = 1\ atm$), the molar volume of a gas is $V_m = 22.4\ L/mol$. The volume of $O_2$, $V_{O_2}$, is calculated using the formula $V=n\times V_m$. So, $V_{O_2}=0.083\ mol\times22.4\ L/mol = 1.8592\ L$. Converting to milliliters, $V_{O_2}=1.8592\ L\times1000\ mL/L\approx1860\ mL$.

Answer:

1860 mL