How many molecules are in 643.21 grams of Mg3(PO4)2?
1 answer:
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Answer: 404.04 kJ.
Explanation:
To calculate the moles, we use the equation:
moles of
According to stoichiometry :
2 moles of on burning produces = 1036 kJ
Thus 0.78 moles of on burning produces =
Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.
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