Question

I have two problems that if someone could kindly answer it that would be awesome.

1. How many grams of CO are needed to react with an excess of Fe₂O₃ to produce 225.5g Fe? Fe₂O₃(g) +3CO(g) → 3CO₂(g) + 2Fe(s)

2.Solid sodium reacts violently with water producing heat, hydrogen gas and sodium hydroxide. How many molecules of hydrogen gas are produced when 65.4g of sodium are added to water? 2Na (s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Answer 1

1. Because there is an excess of Fe2O3 we do not really need to consider that in our calculations.

The first thing you will need to do in this question is convert the grams of Fe we have (225.5g) into the moles. This is because we can use the moles to find the moles of CO to find the grams.

To find the moles, you will need to do: moles = mass/molar mass
So its 225.5g / (molar mass of Fe) = moles of Fe

Once you have the moles, you can then look at the stoichiometry to find the moles of CO. Its a 3:2 ratio and 1 mole of every substance would be the same. So your answer will be TWO lots of Fe, and you need THREE lots of CO. So you will need to divide your answer by 2 to get ONE lot of moles and then times it by THREE to get the actual moles of CO.
Moles of Fe / 2 x 3 = moles of CO

You will then need to convert the moles of CO into grams (aka. mass) by doing: moles x molar mass of CO = mass (g)

Do this and you will end up with your answer for number 1. I need to go now I could help you with the second one later if no one else does before me :) Good luck

Answer 2

Thank you so much ;D

Answer 3

2. I guess there is excess of water in the reaction.

First, calculate amount of substance of Na, also n(Na)
Then you see, amount of substance of H2 is one half of n(Na).
Multiply n(H2) with Avogadro number.