Question

common polyatomic ions
chemical name chemical formula chemical name chemical formula
acetate $c_{2}h_{3}o_{2}^{-}$ nitrite $no_{2}^{-}$
carbonate $co_{3}^{2 - }$ ammonium $nh_{4}^{+}$
hypocarbonite $co^{2 - }$ cyanide $cn^{-}$
hydrogen carbonate (bicarbonate) $hco_{3}^{-}$ hydroxide $oh^{-}$
chlorite $clo_{2}^{-}$ peroxide $o_{2}^{2 - }$
hypochlorite $clo^{-}$ phosphate $po_{4}^{3 - }$
chlorate $clo_{3}^{-}$ hydrogen phosphate $hpo_{4}^{2 - }$
perchlorate $clo_{4}^{-}$ dihydrogen phosphate $h_{2}po_{4}^{-}$
hydronium $h_{3}o^{+}$ sulfate $so_{4}^{2 - }$
permanganate $mno_{4}^{-}$ hydrogen sulfate (bisulfate) $hso_{4}^{-}$
16
select the correct answer.
sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. if you have 379.4 grams of calcium chloride and an excess of sodium phosphate, how much calcium phosphate can you make? use the periodic table and the polyatomic ion resource.
a. 353.5 g
b. 379.4 g
c. 402.2 g
d. 863.4 g
e. 1060 g

Explanation:

Step1: Write the balanced chemical equation

$2Na_3PO_4 + 3CaCl_2=6NaCl + Ca_3(PO_4)_2$

Step2: Calculate molar - mass of $CaCl_2$ and $Ca_3(PO_4)_2$

Molar mass of $CaCl_2$: $M_{CaCl_2}=40.08+(2\times35.45)=110.98\ g/mol$
Molar mass of $Ca_3(PO_4)_2$: $M_{Ca_3(PO_4)_2}=(3\times40.08)+(2\times(30.97 + 4\times16.00)) = 310.18\ g/mol$

Step3: Calculate moles of $CaCl_2$

$n_{CaCl_2}=\frac{m_{CaCl_2}}{M_{CaCl_2}}=\frac{379.4\ g}{110.98\ g/mol}=3.42\ mol$

Step4: Determine mole - ratio from the balanced equation

The mole - ratio of $CaCl_2$ to $Ca_3(PO_4)_2$ is $3:1$. So, $n_{Ca_3(PO_4)_2}=\frac{1}{3}n_{CaCl_2}$
$n_{Ca_3(PO_4)_2}=\frac{1}{3}\times3.42\ mol = 1.14\ mol$

Step5: Calculate mass of $Ca_3(PO_4)_2$

$m_{Ca_3(PO_4)_2}=n_{Ca_3(PO_4)_2}\times M_{Ca_3(PO_4)_2}$
$m_{Ca_3(PO_4)_2}=1.14\ mol\times310.18\ g/mol = 353.6\ g\approx353.5\ g$

Answer:

A. 353.5 g