# chemistry

Magnesium nitride is formed in the reaction of magnesium metal with nitrogen gas in this reaction: 3 Mg(s) + N2(g) --> Mg3N2(s)
How many grams of product are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg(s)?

i havent been able to figure this one out, can someone help me?

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1. First thing first: write and balance the euqation. In this case, it is already balance. Next, given the moles of the reactants, you should be able to identify which one of them is in excess and which one is limiting agent. The ratio of Mg to N2 is 3:1, meaning if you have 2 mol of N2, you should have had 6 mol of Mg. Hence mg is in excess (containing 8 instead of 6)..
Thirdly, consider the molar ratio of the entire reaction, basing from the limiting reatant (N2). From the equation N2 to Mg3N2 are in the ratio of 1 as to 1. So The product also has 2 mol.
Fourthly, use the formula mass=mol*molecular mass. Use your period table to find the total mass of the product and multiply it with the 0.2
Got it?

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2. sorry .-. I don't understand

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3. This is a limiting reagent (LR) problem; you know that because amounts are given for both reactants.

3Mg(s) + N2(g) --> Mg3N2(s)
The long long way to do this.

mols Mg = 8 from the problem.
mols N2 = 2 from the problem.

Using the coefficients in the balanced equation, convert mols Mg to mols Mg3N2.

Do the same and convert mols N2 to mols Mg3N2.

It is likely that these two values will not be the same; the correct value in LR problem is ALWAYS the smaller value and the reagent responsible for that value is called the LR.

Now use the smaller value and convert to grams. g = mols Mg3N2 x molar mass Mg3N2 = ?

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