common polyatomic ions\nchemical name chemical formula chemical name chemical formula\nacetate…

common polyatomic ions\nchemical name chemical formula chemical name chemical formula\nacetate $c_{2}h_{3}o_{2}^{-}$ nitrite $no_{2}^{-}$\ncarbonate $co_{3}^{2 - }$ ammonium $nh_{4}^{+}$\nhypocarbonite $co^{2 - }$ cyanide $cn^{-}$\nhydrogen carbonate (bicarbonate) $hco_{3}^{-}$ hydroxide $oh^{-}$\nchlorite $clo_{2}^{-}$ peroxide $o_{2}^{2 - }$\nhypochlorite $clo^{-}$ phosphate $po_{4}^{3 - }$\nchlorate $clo_{3}^{-}$ hydrogen phosphate $hpo_{4}^{2 - }$\nperchlorate $clo_{4}^{-}$ dihydrogen phosphate $h_{2}po_{4}^{-}$\nhydronium $h_{3}o^{+}$ sulfate $so_{4}^{2 - }$\npermanganate $mno_{4}^{-}$ hydrogen sulfate (bisulfate) $hso_{4}^{-}$\n16\nselect the correct answer.\nsodium 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.\na. 353.5 g\nb. 379.4 g\nc. 402.2 g\nd. 863.4 g\ne. 1060 g

common polyatomic ions\nchemical name chemical formula chemical name chemical formula\nacetate $c_{2}h_{3}o_{2}^{-}$ nitrite $no_{2}^{-}$\ncarbonate $co_{3}^{2 - }$ ammonium $nh_{4}^{+}$\nhypocarbonite $co^{2 - }$ cyanide $cn^{-}$\nhydrogen carbonate (bicarbonate) $hco_{3}^{-}$ hydroxide $oh^{-}$\nchlorite $clo_{2}^{-}$ peroxide $o_{2}^{2 - }$\nhypochlorite $clo^{-}$ phosphate $po_{4}^{3 - }$\nchlorate $clo_{3}^{-}$ hydrogen phosphate $hpo_{4}^{2 - }$\nperchlorate $clo_{4}^{-}$ dihydrogen phosphate $h_{2}po_{4}^{-}$\nhydronium $h_{3}o^{+}$ sulfate $so_{4}^{2 - }$\npermanganate $mno_{4}^{-}$ hydrogen sulfate (bisulfate) $hso_{4}^{-}$\n16\nselect the correct answer.\nsodium 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.\na. 353.5 g\nb. 379.4 g\nc. 402.2 g\nd. 863.4 g\ne. 1060 g

Answer

Answer:

A. 353.5 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 $Ca_3(PO_4)_2$

$m_{Ca_3(PO_4)2}=n{Ca_3(PO_4)2}\times M{Ca_3(PO_4)2}$ $m{Ca_3(PO_4)_2}=1.14\ mol\times310.18\ g/mol = 353.6\ g\approx353.5\ g$