{"id":43384,"date":"2025-06-29T19:15:06","date_gmt":"2025-06-29T19:15:06","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=43384"},"modified":"2025-06-29T19:15:07","modified_gmt":"2025-06-29T19:15:07","slug":"if-you-added-an-excess-of-sodium-phosphate-dodecahydrate-to-2-122-g-of-barium-chloride-dihydrate-how-many-grams-of-barium-phosphate-could-you-synthesize","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/if-you-added-an-excess-of-sodium-phosphate-dodecahydrate-to-2-122-g-of-barium-chloride-dihydrate-how-many-grams-of-barium-phosphate-could-you-synthesize\/","title":{"rendered":"If you added an excess of sodium phosphate dodecahydrate to 2.122 g of barium chloride dihydrate, how many grams of barium phosphate could you synthesize"},"content":{"rendered":"\n

If you added an excess of sodium phosphate dodecahydrate to 2.122 g of barium chloride dihydrate, how many grams of barium phosphate could you synthesize? Note: Report your answer to three decimal places<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To solve this problem, we need to determine how many grams of barium phosphate (Ba3(PO4)2\\text{Ba}_3\\text{(PO}_4)_2Ba3\u200b(PO4\u200b)2\u200b) can be synthesized when an excess of sodium phosphate dodecahydrate (Na3PO4\u22c512H2O\\text{Na}_3\\text{PO}_4 \\cdot 12\\text{H}_2\\text{O}Na3\u200bPO4\u200b\u22c512H2\u200bO) is added to 2.122 grams of barium chloride dihydrate (BaCl2\u22c52H2O\\text{BaCl}_2 \\cdot 2\\text{H}_2\\text{O}BaCl2\u200b\u22c52H2\u200bO).<\/p>\n\n\n\n

Step 1: Write the balanced chemical equation<\/h3>\n\n\n\n

The reaction between barium chloride and sodium phosphate can be represented as:3BaCl2+2Na3PO4\u2192Ba3(PO4)2+6NaCl3 \\text{BaCl}_2 + 2 \\text{Na}_3\\text{PO}_4 \\rightarrow \\text{Ba}_3\\text{(PO}_4)_2 + 6 \\text{NaCl}3BaCl2\u200b+2Na3\u200bPO4\u200b\u2192Ba3\u200b(PO4\u200b)2\u200b+6NaCl<\/p>\n\n\n\n

Step 2: Calculate the molar mass of BaCl2\u22c52H2O\\text{BaCl}_2 \\cdot 2\\text{H}_2\\text{O}BaCl2\u200b\u22c52H2\u200bO<\/h3>\n\n\n\n

First, we calculate the molar mass of barium chloride dihydrate:<\/p>\n\n\n\n

    \n
  • Barium (Ba): 137.33 g\/mol<\/li>\n\n\n\n
  • Chlorine (Cl): 35.45 g\/mol (x2 for BaCl2\\text{BaCl}_2BaCl2\u200b)<\/li>\n\n\n\n
  • Water (H\u2082O): 18.015 g\/mol (x2 for 2H2O2\\text{H}_2\\text{O}2H2\u200bO)<\/li>\n<\/ul>\n\n\n\n

    So, the molar mass of BaCl2\u22c52H2O\\text{BaCl}_2 \\cdot 2\\text{H}_2\\text{O}BaCl2\u200b\u22c52H2\u200bO is:137.33+2(35.45)+2(18.015)=137.33+70.90+36.03=244.26\u2009g\/mol137.33 + 2(35.45) + 2(18.015) = 137.33 + 70.90 + 36.03 = 244.26 \\, \\text{g\/mol}137.33+2(35.45)+2(18.015)=137.33+70.90+36.03=244.26g\/mol<\/p>\n\n\n\n

    Step 3: Convert mass of barium chloride to moles<\/h3>\n\n\n\n

    Next, we convert the mass of barium chloride dihydrate (2.122 g) to moles:Moles of BaCl2\u22c52H2O=2.122\u2009g244.26\u2009g\/mol=0.00868\u2009mol\\text{Moles of } \\text{BaCl}_2 \\cdot 2\\text{H}_2\\text{O} = \\frac{2.122 \\, \\text{g}}{244.26 \\, \\text{g\/mol}} = 0.00868 \\, \\text{mol}Moles of BaCl2\u200b\u22c52H2\u200bO=244.26g\/mol2.122g\u200b=0.00868mol<\/p>\n\n\n\n

    Step 4: Use the stoichiometry of the balanced equation<\/h3>\n\n\n\n

    From the balanced equation, we know that 3 moles of BaCl2\\text{BaCl}_2BaCl2\u200b react with 2 moles of Na3PO4\\text{Na}_3\\text{PO}_4Na3\u200bPO4\u200b to produce 1 mole of Ba3(PO4)2\\text{Ba}_3\\text{(PO}_4)_2Ba3\u200b(PO4\u200b)2\u200b.<\/p>\n\n\n\n

    The molar ratio of BaCl2\\text{BaCl}_2BaCl2\u200b to Ba3(PO4)2\\text{Ba}_3\\text{(PO}_4)_2Ba3\u200b(PO4\u200b)2\u200b is 3:1. Therefore, the moles of Ba3(PO4)2\\text{Ba}_3\\text{(PO}_4)_2Ba3\u200b(PO4\u200b)2\u200b produced is:Moles of Ba3(PO4)2=0.00868\u2009mol\u2009BaCl23=0.00289\u2009mol\u2009Ba3(PO4)2\\text{Moles of } \\text{Ba}_3\\text{(PO}_4)_2 = \\frac{0.00868 \\, \\text{mol} \\, \\text{BaCl}_2}{3} = 0.00289 \\, \\text{mol} \\, \\text{Ba}_3\\text{(PO}_4)_2Moles of Ba3\u200b(PO4\u200b)2\u200b=30.00868molBaCl2\u200b\u200b=0.00289molBa3\u200b(PO4\u200b)2\u200b<\/p>\n\n\n\n

    Step 5: Calculate the mass of Ba3(PO4)2\\text{Ba}_3\\text{(PO}_4)_2Ba3\u200b(PO4\u200b)2\u200b<\/h3>\n\n\n\n

    Next, we calculate the molar mass of barium phosphate:<\/p>\n\n\n\n

      \n
    • Barium (Ba): 137.33 g\/mol (x3 for Ba3\\text{Ba}_3Ba3\u200b)<\/li>\n\n\n\n
    • Phosphorus (P): 30.974 g\/mol (x2 for PO4\\text{PO}_4PO4\u200b)<\/li>\n\n\n\n
    • Oxygen (O): 16.00 g\/mol (x8 for PO4\\text{PO}_4PO4\u200b)<\/li>\n<\/ul>\n\n\n\n

      The molar mass of Ba3(PO4)2\\text{Ba}_3\\text{(PO}_4)_2Ba3\u200b(PO4\u200b)2\u200b is:3(137.33)+2(30.974)+8(16.00)=411.99+61.948+128.00=601.938\u2009g\/mol3(137.33) + 2(30.974) + 8(16.00) = 411.99 + 61.948 + 128.00 = 601.938 \\, \\text{g\/mol}3(137.33)+2(30.974)+8(16.00)=411.99+61.948+128.00=601.938g\/mol<\/p>\n\n\n\n

      Now, we calculate the mass of barium phosphate produced:Mass of Ba3(PO4)2=0.00289\u2009mol\u00d7601.938\u2009g\/mol=1.738\u2009g\\text{Mass of } \\text{Ba}_3\\text{(PO}_4)_2 = 0.00289 \\, \\text{mol} \\times 601.938 \\, \\text{g\/mol} = 1.738 \\, \\text{g}Mass of Ba3\u200b(PO4\u200b)2\u200b=0.00289mol\u00d7601.938g\/mol=1.738g<\/p>\n\n\n\n

      Final Answer:<\/h3>\n\n\n\n

      The amount of barium phosphate that could be synthesized is 1.738 g<\/strong>.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      If you added an excess of sodium phosphate dodecahydrate to 2.122 g of barium chloride dihydrate, how many grams of barium phosphate could you synthesize? Note: Report your answer to three decimal places The Correct Answer and Explanation is: To solve this problem, we need to determine how many grams of barium phosphate (Ba3(PO4)2\\text{Ba}_3\\text{(PO}_4)_2Ba3\u200b(PO4\u200b)2\u200b) can […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-43384","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43384","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=43384"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43384\/revisions"}],"predecessor-version":[{"id":43386,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43384\/revisions\/43386"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43384"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43384"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43384"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}