The mass of butane that can burn excess O2 to produce 18.4kj of heat energy is 6.47g. That is option B
The correct coefficient for the balanced equation is= 2,1 --> 4,1. That is option D
Calculation of heat energy
The molar mass of butane = 58.12 g/mol
For the balanced equation, molar mass of butane
= 2×58.12 = 116.24g/mol
If 116.24g = 330KJ of heat
X g = 18.4kj
Make X the subject of formula,
xg = 116.24 × 18.4/330
xg= 2138.816/330
xg= 6.47g
Therefore, The mass of butane that can burn excess O2 to produce 18.4kj of heat energy is 6.47g.
The balance equation from the question is
2K2Cr2O7 + SnCl4 ---> 4KCl + Sn(Cr2O7)2
Which has the coefficient of 2,1 --> 4,1
Learn more about balanced equation here:
https://brainly.com/question/26227625
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