A student burned 0.250 g of magnesium metal in an open crucible.Oxygen from the air combined with the metal forming magnesium ox
1 answer:
Answer:
0.171 grams of oxygen combined with the metal.
Explanation:
Mass of magnesium oxide produced = 0.421 g
Mass of magnesium metal used = 0.250 g
Suppose reaction completely converts the magnesium into magnesium oxide.
Mass of oxygen combined with magnesium = m
0.421 g = 0.250 g + m
m = 0.421 g - 0.250 g = 0.171 g
0.171 grams of oxygen combined with the metal.
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Answer:
Explanation:
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In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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