There is a shorter way of doing this but I like the long way.
Convert moles each to the product, then take the smaller product amount as coming from the limiting reagent.
2 moles N2 x (2 moles NH3/2 moles N2) = 2 moles NH3.(The fractions comes from the coefficients.)
5 moles H2 x (2 moles NH3/3 moles H2) = 3.33 moles NH3.
The smaller number is N2, that is the limiting reagent.
The others are done the same way. For excess reagent, use the limiting reagent, convert to the excess reagent with the appropriate fraction to determine how much is used in the reaction, subtract that from the moles initially there to determine the amount that did not react.
ammonia is produced by sythesizing nitrogen and hydrogen gas. if i have 2 mols of nitrogen gas and 5 mols of hydrogen gas, which reagent will run out first??
a) if i have 0.0345 mols of nitrogen gas and 0.0537 mols of hydrogen gas, which reagent will run out first? how many mols of excess reagent will have left over?
b) if i have 24g of nitrogen gas and 24g of hydrogen gas, which reagent will run out first?
2 answers
hey can u show me the shorter way to do plzzz and thx for the answer... :)
1 answer