I think that the empirical formula is mgcl2
What is the second quantum number of a 3p3 electron in phosphorus 1s22s22p63s23p3
1 answer:
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Answer:
0.677 moles
Explanation:
Take the atomic mass of K = 39.1, O =16.0, P = 31.0
no. of moles = mass / molar mass
no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)
= 0.02256 mol
From the equation, the mole ratio of KOH : K3PO4 = 3 :1,
meaning every 3 moles of KOH used, produces 1 mole of K3PO4.
So, using this ratio, let the no. of moles of KOH required to be y.
y = 0.02256 x3
y = 0.0677 mol
If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.