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Pavel [41]
3 years ago
10

What volume (in mililiters) of oxygen gas is required to react with 4.03 g of my at stp

Chemistry
1 answer:
goldenfox [79]3 years ago
4 0

You must use 1880 mL of O₂ to react with 4.03 g Mg.

A_r: 24.305

         2Mg + O₂ ⟶ 2MgO

<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg

<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂

STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.

<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880  mL

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