If 62.7 g N2 react with 23.8 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced, and how many grams of the excess reactant will be left over? Unbalanced equation: N2 + H2 → NH3
@computergeek123 The question seems to be perfect for the Chemistry Group. Why don't you post it there? This group is meant for asking questions related to Mathematics only. http://openstudy.com/terms-and-conditions
N2+H2-> NH3 N2+3H2->2NH3 ^ ^23.8g 62.7g no. of moles = mass/molar mass calculate no of moles of N2 and H2 then by conservation, we can get the no of moles of NH3 products then by no. of moles = mass/molar mass, we can get mass of product
First, please post in the right group next time Next, I don't feel like repeating what CoCoTsoi has mentioned above. Just one thing, what you need is the mass of excess reactant. So after you find the number of mole of reactants, you'd probably be able to determine the excess reactant. Then you should minus it with the theoretical amount you need for the reactant. You'll get the no of mole of unreacted reactant. After that by mole = mass / molar mass, you can find the mass by multiplying the no of mole of unreacted reactants by its molar mass.
To be more practical, N2 + H2 → NH3 Balance the equation N2 + 3H2 → 2NH3 no. of mole of N2 = 62.7 / (14.0x2) = 2.24 moles no of mole of H2 = 23.8 / (1.0x2) = 11.9 moles No of mole of H2 need = 3 x no. of mole of N2 reacted = 3x2.24 = 6.72moles Therefore H2 is in excess No of mole of unreacted H2 = 11.9 - 6.72 = 5.18 moles Mass of unreacted H2 = mole x molar mass = 5.18 x (1.0x2) Can you finish the remaining part?
No, I don't get this question like at all. But the part you explained helped a little bit.
What part you don't understand?
How to finish the equation.
use the equation mole = mass/ molar mass Actually, i've explained the way to do it.. Which part you don't understand?
so do i multiply 5.18 x 1.0x2??
yes, you need to
10.36??
i suppose it should be like that
okay what next?
That is the answer already...
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