How many grams of ammonia (NH3) are produced when 15.0 g of hydrogen reacts with an excess of nitrogen?

Question

see above

Answer 1

Collect the data from the question.

You have 15.0g H2

Excess Nitrogen (which means all of the hydrogen will be used up!)

get your periodic table and find the molecular masses (for H2 1.008+1.008=2.016g/mol)

N

you need to set up your equation, both of them.

H2 + N2 --> NH3 blanace it (very important)

3H2 + N2 --> 2NH3 (2 atoms of N on reactants side, need 2 on the products side. now there are 6 hydrogens on the products side, and 2 on the reactants, multiply 2 by 3 for 6 on each side.)

second equation:

start with the hydrogen and multiply the grams by the molar mass of hydrogen (which i assume is a gas). now you have moles of hydrogen you need to convert this to moles of NH3 (this is where your molecular formula comes in.)  you have moles of hydrogen (3), this will go on the top to cancel the moles. divide by moles of NH3 (2). Now you have mols to moles of NH3. Now just divide by the molar mass of NH3.

15.0g H2/2.016g/mol H2=7.44mol H2 x 3mol H2/2mol NH3= 11.16mol NH3 x 1/17.02g/mol NH3= your answer

i would suggest to do this equation all at once on your calculator so you do not miss any significant figures.

Answer 2

Okay so first right down what you know:

H2 = 15.0g

N2 = excess

NH3 = ?

So because they told you ur N2 is excess, therefore your hydrogen would be ur limiting reagent. So firstly you MUST balance your equation:

H2 + N2 -> NH3   (unbalanced)

3H2 + N2 -> 2NH3 (balanced)

Now, find out the mole of hydrogen:

15.0g / 2.02 g/mol    = 7.4257... mol    

Since you know that the coefficient of H2 is 3, you have to divide that from your 7.4257..mol inorder to find out how many mole is in ONE H2:

7.4257..mol/ 3  = 2.47...mol   (that's your limit)

From there, you can multiply by 2 (because that's the coefficient of NH3) to find out the total mole of NH3:

2.47...mol x 2 = 4.950... mol

Then you can find the mass of NH3 by using the equation mole(molaar mass) = mass

4.950mol x 17.04 g/mol = 84.4 g

Therefore, the mass of NH3 is 84.4g.