the chemical equation below shows the burning of magnesium (mg) with oxygen (o₂) to form magnesium oxide…

the chemical equation below shows the burning of magnesium (mg) with oxygen (o₂) to form magnesium oxide (mgo).\n2mg + o₂ → 2mgo\nthe molar mass of o₂ is 32.0 g/mol. what mass, in grams, of o₂ is required to react completely with 4.00 mol of mg?\n2.00 grams\n64.0 grams\n128 grams\n256 grams

the chemical equation below shows the burning of magnesium (mg) with oxygen (o₂) to form magnesium oxide (mgo).\n2mg + o₂ → 2mgo\nthe molar mass of o₂ is 32.0 g/mol. what mass, in grams, of o₂ is required to react completely with 4.00 mol of mg?\n2.00 grams\n64.0 grams\n128 grams\n256 grams

Answer

Explanation:

Step1: Determine the mole ratio of (Mg) to (O_2)

From the balanced chemical equation (2Mg + O_2\rightarrow2MgO), the mole ratio of (Mg) to (O_2) is (2:1).

Step2: Calculate the moles of (O_2) needed

If there are (n(Mg) = 4.00\space mol) of (Mg), using the mole ratio (\frac{n(O_2)}{n(Mg)}=\frac{1}{2}), then (n(O_2)=\frac{1}{2}\times n(Mg)). Substituting (n(Mg) = 4.00\space mol), we get (n(O_2)=\frac{1}{2}\times4.00\space mol = 2.00\space mol).

Step3: Calculate the mass of (O_2)

Using the formula (m = n\times M), where (M = 32.0\space g/mol) (molar mass of (O_2)) and (n = 2.00\space mol) (moles of (O_2)). Then (m(O_2)=n(O_2)\times M(O_2)=2.00\space mol\times32.0\space g/mol) [m(O_2)=64.0\space g]

Answer:

64.0 grams