Given:
2 Mg + O2 → 2 MgO
So,
<span>(25.0 g of Mg / 24.3051 g/mole) x (1/2) x (22.4 L/mole) = 11.5 L of Oxygen will react with 25 grams of magnesium metal.</span>
The statement which is true is
Fluorine is more reactive than nitrogen because fluorine needs only one electron to fill its outermost shell.
<u><em>Explanation</em></u>
Fluorine has electron configuration of 1S²2S²2P⁵ while nitrogen has 1S²2S²2P³ electron configuration.
The 2P sub shell for nitrogen is half filled therefore it is sable than fluorine.
since p orbital can hold a maximum of 6 electrons ,Fluorine requires 1 electron to completely fill it's 2P sub shell which make it more reactive than nitrogen.
Withdrawal is the answer to your question.
The formula of Iron(III) oxide is Fe2O3
In order to calculate the mass of iron in a given sample of iron(III) oxide, we must first know the mass percentage of iron in iron(III) oxide. This is calculated by:
[mass of iron in one mole of iron(III) oxide/ mass of one mole of iron(III) oxide] * 100
= [(moles of iron * Mr of iron) / (moles of Iron * Mr of Iron + moles of Oxygen * Mr of Oxygen)] * 100
= [(2 * 56) / (2 * 56 + 3 * 16)] * 100
= (112 / 160) * 100
= 70%
Thus, in a 100g sample, the weight of iron will be:
100 * 70%
= 70 grams
