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
Stacey Warren - Expert brainly.com
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@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
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.
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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?