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notsponge [240]
3 years ago
14

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 m

L 3710 mL 45,100 mL
Chemistry
2 answers:
charle [14.2K]3 years ago
4 0
<span>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</span>
step 4: multiply 0.0829mol O₂ by 22.4L/mol to get 1.857L
answer is 1860 mL
Sholpan [36]3 years ago
4 0

Answer: 1860 mL

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\times 24.305=48.61g of Magnesium reacts with 22.4 L of oxygen gas STP.

4.03g of  Magnesium reacts with=\frac{22.4L}{48.61g}\times 4.03g=1.86L=1860ml of oxygen gas STP.

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See figure 1

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