Question

question 5 (1 point)
how many equivalents of $\ce{po4^-3}$ are there in 6.00 mol of $\ce{mg3(po4)2}$?
first we have to convert moles into equivalents
$\ce{mg3(po4)2 \
ightarrow 3 mg^+2 + 2 po4^-3}$
1 mol = 1 equivalents
2 eq
(6.00 mol mp)(---------)= 3 eq $\ce{po4^-3}$
4 mol
a. 250. b. 100.0 c. 2 d. 30.00 e. 6 f. 24
g. 3 h. 36 i. 18.0 j. 1 k. $6.00 \times 10^{-3}$ l. $3.60 \times 10^{-2}$
m. $6.00 \times 10^{-2}$ n. 5 o. $3.0 \times 10^{-2}$ p. $1.80 \times 10^{-2}$
q. $7.65 \times 10^{-10}$ r. 2.63 s. 7.89 t. 5.26 u. 95.21
v. $10^{12}$ w. $10^9$ x. $10^6$ y. $10^3$ z. 105.21

Explanation:

Step1: Determine equivalents per mole of \(Mg_3(PO_4)_2\)

From the dissociation \(Mg_3(PO_4)_2
ightarrow3Mg^{ + 2}+2PO_4^{-3}\), for \(PO_4^{-3}\), \(1\) mole of \(Mg_3(PO_4)_2\) gives \(2\) moles of \(PO_4^{-3}\). Since the charge on \(PO_4^{-3}\) is \(3-\), the number of equivalents per mole of \(PO_4^{-3}\) is \(3\). So equivalents per mole of \(Mg_3(PO_4)_2\) for \(PO_4^{-3}\): \(2\times3 = 6\) (so \(1\) is \(6\)).

Step2: Set up the conversion factor

The conversion factor is \(\frac{6\space eq}{1\space mol}\) (so \(2\) is \(6\), \(4\) is \(1\)).

Step3: Calculate equivalents

Using the formula \(Equivalents=(moles\space of\space Mg_3(PO_4)_2)\times\frac{equivalents}{moles}\), \((6.00\space mol\space Mg_3(PO_4)_2)\times\frac{6\space eq}{1\space mol}=36\space eq\) (so \(3\) is \(36\)).

Answer:

1. \(6\)
2. \(6\)
3. \(36\)
4. \(1\)