If you have 9 moles of CaCl2 and an excess of Na3PO4, how many moles of NaCl can be produced according to the following reaction?CaCl2 + Na3PO4 → Ca3(PO4)2 + NaCl

• A. 3 moles
• B. 18 moles
• C. 27 moles
• D. 9 moles

Answer: (B)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction. Start by balancing the number of atoms for each element on both sides. The unbalanced equation is: \(\mathrm{CaCl_2 + Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + NaCl}\).
2. Balance calcium (Ca) atoms first. Since there are 3 Ca atoms in \(\mathrm{Ca_3(PO_4)_2}\), place a coefficient of 3 in front of \(\mathrm{CaCl_2}\): \$3\ \mathrm{CaCl_2 + Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + NaCl}$.
3. Balance phosphate (\(\mathrm{PO_4}\)) groups next. There are 2 phosphate groups in \(\mathrm{Ca_3(PO_4)_2}\), so place a coefficient of 2 in front of \(\mathrm{Na_3PO_4}\): \$3\ \mathrm{CaCl_2 + 2\ Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + NaCl}$.
4. Balance sodium (Na) atoms. On the left, there are \$2 \times 3 = 6\( Na atoms from \)2\ \mathrm{Na_3PO_4}\(. Therefore, place a coefficient of 6 in front of \)\mathrm{NaCl}\( on the right: \)3\ \mathrm{CaCl_2 + 2\ Na_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6\ NaCl}$.
5. Use the mole ratio from the balanced equation to find moles of \(\mathrm{NaCl}\) produced. For every 3 moles of \(\mathrm{CaCl_2}\), 6 moles of \(\mathrm{NaCl}\) are produced. Set up the proportion: \(\frac{6\ \mathrm{mol\ NaCl}}{3\ \mathrm{mol\ CaCl_2}} = \frac{x\ \mathrm{mol\ NaCl}}{9\ \mathrm{mol\ CaCl_2}}\). Solve for \(x\) to find the moles of \(\mathrm{NaCl}\) produced.