Question

Consider this reaction. What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP? 1860 mL 2880 mL 3710 mL 45,100 mL

Answer 1

Answer: 1860 mL

Step-by-step explanation:

`2Mg+O_2\rightarrow 2MgO`

According to Avogadro's law , 1 mole of every gas occupies 22.4 L at standard temperature and pressure and 1 mole of every substance weighs equal to its molar mass.

`2moles=2* 24.305=48.61g` of Magnesium reacts with 22.4 L of oxygen gas STP.

4.03g of Magnesium reacts with=
`(22.4L)/(48.61g)* 4.03g=1.86L=1860ml` of oxygen gas STP.

Answer 2

Step 1: find number of moles of Mg(divide 4.03 by its molor mass)
step 2:find number of moles in O2
step 3: remember STP is 22.4 L = 1 mole
step 4: multiply 0.0829mol O₂ by 22.4L/mol to get 1.857L
answer is 1860 mL