Answer
a. $4.5∗10^{−3}$ mol $P_4O_6$
b. $3.2∗10^{−3}$ mol $Ca_3(PO_4)_2$
c. $7.0∗10^{−3}$ mol $Na_2HPO_4$$
Work Step by Step
a.
$1.00\text{ grams $P_4O_6$}∗\frac{1 \text{ mol $P4O6$}}{219.89 \text{ grams $P_4O_6$}}=4.5∗10^{−3} \text{ mol $P_4O_6$}$
b.
$1.00\text{ grams $Ca_3(PO_4)_2$}∗\frac{1 \text{ mol $Ca_3(PO_4)_2$}}{310.17 \text{ grams $Ca_3(PO_4)_2$}}=3.2∗10^{−3}\text{ mol $Ca_3(PO_4)_2$ }$
c.
$1.00 \text{ grams $Na_2HPO_4$}∗\frac{1 \text{ mol $Na_2HPO_4$}}{141.96\text{ grams $Na2HPO4$}}=7.0∗10^{−3} \text{ mol $Na_2HPO_4$}$
