When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, MgO. The other is the product of the reaction of Mg with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a 6.3-g Mg ribbon reacts with 2.57 g NH31g2 and the reaction goes to completion, which component is the limiting reactant? What mass of H21g2 is formed in the reaction?

Question

Accepted Answer

hey everyone in this example, we're told that alkali metals are highly reactive and react violently with water. We are dealing with a reaction of sodium metal with water to produce sodium hydroxide and hydrogen gas. And we're told that due to the XR thermic nature of this reaction, in high flammability of hydrogen gas flames are produced. We are assuming that this reaction is carried out in the absence of oxygen gas. And we need to determine how much hydrogen gas is produced when 2.55 g of a piece of sodium metal reacts with 5.15 g of liquid water. So what this question is asking for is for our theoretical yield of our product, hydrogen gas produced from the following masses of these two reactant water and sodium metal. So what we're gonna do is find our theoretical mass of hydrogen gas from our first re agent, which is our sodium metal. And then we also want to find our theoretical mass of hydrogen from our second given re agent water. However, before we can go ahead and get into these calculations, we want to write out our balanced equation. So we have sodium metal according to the question reacting with water and it's going to produce sodium hydroxide dot sodium hydroxide, N A. O. H. And hydrogen gas. We want to make sure that this reaction is written out with a balanced with balanced coefficients. And so we want to go ahead and add the following coefficients so that we have a total of two in front of our water. This will then have to balance out with our oxygen. So we'll add a two in front of our sodium hydroxide. And then to balance out our sodium will add a two in front of that. And so now our equation is balanced. We also want to go ahead and write the phases of each of our re agents. So for sodium this is going to be a solid metal. For water, this is liquid for sodium hydroxide, this is an Aquarius reactant. And then we're going to have our hydrogen gas with the gas label here. So now that we have our balanced reaction fully written out, we're going to also make note of the fact that we need our ratio our molar ratio between our re agent. So for the first part sodium to our product hydrogen gas. And we're going to get this ratio from the coefficients and our balanced equation. So looking at the coefficient in front of sodium, We have a coefficient of two. And then for the coefficient in front of hydrogen gas, we have a coefficient of one. So we have a 2-1 molar ratio here. This is important and we'll use this in our calculations. We also want to make note of this for the second part of our calculation, which is our ratio between water And sorry about that. So between water and our hydrogen gas and looking at our coefficient in front of water, we have a coefficient of two. So we would have a 2-1 ratio between hydrogen gas and water. So beginning with our first part of our calculation, our theoretical mass of hydrogen gas produced from sodium, our first reactant. We're gonna start off with the information, they tell us which is 2.55 g of sodium metal. And we want to go from grams to moles of sodium. So we're going to recall our molar mass of sodium from our periodic table for one atom of sodium, which we'll see is 22.99 g for one mole of sodium. Now we're gonna go ahead and move into moles of sodium And we want to get two moles of our H2 gas, which is our product and then after moles of H two gas, we want to go from moles of H two gas to find out how many grams or the mass of H two gas that is produced. So, for this first conversion here, we want to plug in that Molar ratio that we took note of here and we said that that ratio is a 2-1 ratio. So we said for two moles of sodium, we have two moles of hydrogen gas that is produced. Now we're able to cancel our units of moles of sodium as well as grams of sodium. And now we want to go ahead and plug in our molar mass for hydrogen gas from the periodic table, which we would get for one mole of hydrogen gas. We have a total of two point oh two g per mole of hydrogen gas and this is our molar mass for two atoms of hydrogen gas. So now we're able to cancel our units of moles of hydrogen gas, leaving us with the mass of hydrogen gas for our final units. And this is going to give us a value equal to 0.112g of hydrogen gas produced. And this is our theoretical Mass that we think we will produce from sodium. So moving on to the second part of our calculation, we are given from the question 5.15 g of water. And we want to go ahead and use the molar mass of water to get two moles of water. So for every atom in water, we would get from our periodic tables, a molar mass equal to 18.02g for one mole of water. Next we want to go ahead and move from moles of water, two moles of hydrogen gas using our molar ratio. And that's what we took note of earlier. So we said, we have for two moles of water, one mole of hydrogen gas, which came from our bounced equation. And just like before we want to find our mass of H2 gas produced. So we want to cancel out moles of H2 gas. So we're going to again recall that for the two atoms of hydrogen in a hydrogen gas, we have a molar mass of two point oh two g for one mole of hydrogen gas. So now we're able to cancel our units moles of hydrogen gas, moles of water and grams of water, leaving us with grams of hydrogen gas as our final unit. And this gives us a value equal to 0.289 g of hydrogen gas produced. So now that we have this comparison of the masses of hydrogen gas produced, we want to go ahead and determine that we have 0.28 or rather I should say 0.112 g of H two, which is less than 0.2 89 g of H two gas. And so because this quantity on the left is less, we would say that therefore the re agent that produced this mass of hydrogen gas, which was our sodium metal. We would say that sodium is the limiting reagent and reaction produces 0.11, two g Of H two. And this will be our final answer to complete this example. So our final answer is this portion here as well as our balanced equation which was asked for us also from this question. So I hope that everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

Verified Solution

Key Concepts

Balancing Chemical Equations

Limiting Reactants

Stoichiometry