Asked by shriya
4NH3 + 5O2 --> 4NO + 6H2O
170g of ammonia (NH3) are allowed to react with 384g of O2.
how many moles of NO is produced? and also how many grams of NO will be produced?
170g of ammonia (NH3) are allowed to react with 384g of O2.
how many moles of NO is produced? and also how many grams of NO will be produced?
Answers
Answered by
bobpursley
figure the moles of NH3 in 170g.
now figure the moles of O2 in 384g
if you have 5/4 more moles of O2 than NH3, then NH3 is the limiting reageant. Moles NO=molesNH3*
if you have less than 5/4 moles of O2, then O2 is limiting. Moles NO = 4/5 moles of O2 reacted.
now figure the moles of O2 in 384g
if you have 5/4 more moles of O2 than NH3, then NH3 is the limiting reageant. Moles NO=molesNH3*
if you have less than 5/4 moles of O2, then O2 is limiting. Moles NO = 4/5 moles of O2 reacted.
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