Convert 5,000 kg NH3 to moles.
Convert 5,000 kg O2 to moles.
Convert 5,000 kg CH4 to moles.
Now convert each of those moles, individually, to moles of the product (HCN). The reactant producing the smallest number of moles HCN will be the limiting reagent. Then convert that many moles HCN to grams. g = moles x molar mass. Post your work if you get stuck.
Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane.
2 NH3(g) + 3 O2(g) + 2 CH4(g) 2 HCN(g) + 6 H2O(g)
If 5.00 x 10^3 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?
i don't understand this at all
can someone explain please
2 answers
Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane.
2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g)
If 5.35 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?
HCN g
H2O g
2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g)
If 5.35 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?
HCN g
H2O g
6 answers