Question

Can I get help on figuring out how much excess reactant will be left over when the reaction is complete?

Answer 1

Find out which reactant is limiting and which is excess, convert limiting to moles using molar mass (from the periodic table) convert to moles of final product using molar ratio from balanced stoichiometric equation, convert to grams of final product using molar mass to find out how much product can be formed from limiting reactant.

Take the mass of product formed, convert to moles using molar mass, convert to moles of excess reagent using molar ratio, convert to grams using molar mass.

Take that number and subtract it from your given mass of excess reactant to find out how much will be left over.

Answer 2

Calculate the mass of ammonia produced when 30.0 g of nitrogen react with 11.4 g of hydrogen.

N2 +3H2 --> 2NH3

I got 36.5 g for NH3 and I know hydrogen is the limiting reactant.

Answer 3

So I would subtract the 64.215 from11.4?

Answer 4

You get 36.4798 g of NH3 produced from 30g of N2

You get 64.07g of NH3 produced from 11.4 g of H2

Thus, N2 is your limiting reactant. You can only make 36.4798g before you run out of material, understand?

So then you take that 36.4798g of NH3, which you understand to be the most NH3 that you can create from the materials you have, and convert it grams of your excess to see how much will be left over: g NH3 -> mol NH3-> mol->H2-> g H2

That should give you 6.490539093 g H2 used in the reaction.

Then subtract that from 11.4 g H2 that you had in the beginning, and your answer is 4.909460907g of H2 left over after the reaction.

11.4g was your given so that's 3 significant figures so your answer rounds to 4.91g H2 leftover.

I hope that's helpful

Answer 5

Thanks!

Answer 6

Of course! Medal me bro