Answer to Question #105005 in General Chemistry for Kaci

Question

If 4.5 L of oxygen reacts with magnesium metal how many atoms of magnesium are produced from this reaction?

Answer 1

Mg+O₂→MgO₂ ;

V(O₂)/V_m = N(Mg)/N_A

• V(O₂) - volume of oxygen;
• V_m - molar volume (V_m=22.4);
• N(Mg) -  atoms of magnesium produced from this reaction;
• N_A - avogadro constant (N_A=6.02*10^23);

N(Mg) = V(O₂)*N_A/V_m = 4.5*6.02*10^23 / 22.4 = 1.209 * 10^23;

Answer: 1.209 * 10^23 atoms of magnesium are produced from this reaction.