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Answer: 16.32 g of as excess reagent are left.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of require = 1 mole of
Thus 0.34 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and
is the excess reagent.
Moles of left = (0.68-0.17) mol = 0.51 mol
Mass of
Thus 16.32 g of as excess reagent are left.
Answer:
Explanation:
Hello!
In this case, when we want to balance chemical reactions such as in this case, the idea is to equal to number of atoms of each element at each side of the equation according to the lay of conservation of mass, just as shown below:
Because we have four phosphorous and ten oxygen atoms at each side.
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