How many grams of oxygen are required for the combustion of 25.5 grams of magnesium? 2Mg (s) + O² (g) - > 2MgO (s)

Question

Zariah Hoffman

Chemistry

26 July, 01:17

How many grams of oxygen are required for the combustion of 25.5 grams of magnesium? 2Mg (s) + O² (g) - > 2MgO (s)

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Janiya · Answer 1

Step 1 : Convert grams of magnesium to moles using n = m/MM.

MM is found in periodic table if you don't know it yet.

So we get : n (Mg) = 25.5 g : 24.305 g/mol. = 1.0491 ... mol.

Step 2: Look at the stoichiometric ratios in the formula. Mg : O = 2:1.

Therefore, the moles of oxygen is half the number of moles of Magnesium.

n (O2) = 0.5 x 1.0491 ... mol = 0.52458 ... mol

Step 3: Now convert moles of Oxygen to grams using m = n x MM. Hey presto you have your answer!

m (O2) = 0.52458 ... x 2 (15.999) = 16.7856 ... grams.

To 3 significant figures this is 16.8grams.

Let me know if you are confused about anything.