Ni (s) + 4 CO (g) Ni(CO)4 (g) E = –154 kJ
4*molar mass CO x (756/154)
Given the reaction: Ni (s) + 4 CO (g) Ni(CO)4 (g) E = –154 kJ, how many kilograms
of CO are required for this reaction to release 756 kJ of energy?
3 answers
hi. can you explain to me please why this solution please? THANK YOU
Ni (s) + 4 CO (g) Ni(CO)4 (g) E = –154 kJ
1 mol CO = 12+16 = 28 g
4 mol CO = 4*28 = 112 g
112 g CO produces 154 kJ heat. You want 756 kJ. You know that will take more than 112 g so the ratio is 756/154 and
112 x (756/154) = ?
OR you can do it by proportion.
(112g/154 kJ) = (x/756) and
x = 112*756/154.
Ask yourself this question. If the rxn produces 154 kJ for 112 g CO, then how many g CO would be need to produce just twice as much heat (that's 154*2 = 308). The answer is obvious. If you want twice the heat you will need twice the fuel. How did you get that answer? By 112 x (308/154) = 112*2 = 224g CO
Or how much CO is needed to produce 3x as much heat. Of course the answer is 3x the amount of fuel. How do you get that?(3*154=462)
112 x (462/154) = 336 (or 3*112).
Therefore, if you don't want twice the amount or 3x the amount but you want 756 kJ, you will need 112g x (756 kJ/154 kJ) = ?g
1 mol CO = 12+16 = 28 g
4 mol CO = 4*28 = 112 g
112 g CO produces 154 kJ heat. You want 756 kJ. You know that will take more than 112 g so the ratio is 756/154 and
112 x (756/154) = ?
OR you can do it by proportion.
(112g/154 kJ) = (x/756) and
x = 112*756/154.
Ask yourself this question. If the rxn produces 154 kJ for 112 g CO, then how many g CO would be need to produce just twice as much heat (that's 154*2 = 308). The answer is obvious. If you want twice the heat you will need twice the fuel. How did you get that answer? By 112 x (308/154) = 112*2 = 224g CO
Or how much CO is needed to produce 3x as much heat. Of course the answer is 3x the amount of fuel. How do you get that?(3*154=462)
112 x (462/154) = 336 (or 3*112).
Therefore, if you don't want twice the amount or 3x the amount but you want 756 kJ, you will need 112g x (756 kJ/154 kJ) = ?g