To find the mass of Na\u2083PO\u2084\u00b712H\u2082O<\/strong> (sodium phosphate dodecahydrate) in the mixture, we use stoichiometry based on the information provided:<\/p>\n\n\n\n Moles of Ba\u2083(PO\u2084)\u2082=0.767 g602.0 g\/mol=0.001275 mol\\text{Moles of Ba\u2083(PO\u2084)\u2082} = \\frac{0.767\\ \\text{g}}{602.0\\ \\text{g\/mol}} = 0.001275\\ \\text{mol}Moles of Ba\u2083(PO\u2084)\u2082=602.0 g\/mol0.767 g\u200b=0.001275 mol<\/p>\n\n\n\n From the balanced chemical equation:3BaCl2+2Na3PO4\u2192Ba3(PO4)2\u2193+6NaCl3BaCl\u2082 + 2Na\u2083PO\u2084 \u2192 Ba\u2083(PO\u2084)\u2082\u2193 + 6NaCl3BaCl2\u200b+2Na3\u200bPO4\u200b\u2192Ba3\u200b(PO4\u200b)2\u200b\u2193+6NaCl<\/p>\n\n\n\n So:<\/p>\n\n\n\n Moles of Na\u2083PO\u2084 used=2\u00d70.001275=0.002550 mol\\text{Moles of Na\u2083PO\u2084 used} = 2 \u00d7 0.001275 = 0.002550\\ \\text{mol}Moles of Na\u2083PO\u2084 used=2\u00d70.001275=0.002550 mol<\/p>\n\n\n\n Mass=0.002550 mol\u00d7380.2 g\/mol=0.9695 g\\text{Mass} = 0.002550\\ \\text{mol} \u00d7 380.2\\ \\text{g\/mol} = 0.9695\\ \\text{g}Mass=0.002550 mol\u00d7380.2 g\/mol=0.9695 g<\/p>\n\n\n\n Rounded to three significant figures:0.968 g\\boxed{0.968\\ \\text{g}}0.968 g\u200b<\/p>\n\n\n\n This problem involves using stoichiometry to determine the mass of a reactant based on the amount of precipitate formed. We are told that barium phosphate, Ba\u2083(PO\u2084)\u2082, precipitates from a reaction between barium chloride and sodium phosphate. We also know the mass of the precipitate formed and that sodium phosphate is the limiting reactant, meaning it determines the amount of product made.<\/p>\n\n\n\n To begin, we convert the given mass of barium phosphate to moles using its molar mass. This gives us the amount of product in moles. According to the balanced equation, two moles of sodium phosphate react to form one mole of barium phosphate. So, we multiply the moles of Ba\u2083(PO\u2084)\u2082 by 2 to find the moles of sodium phosphate that were used in the reaction.<\/p>\n\n\n\n After calculating the moles of sodium phosphate, we convert it back to mass. Since the sodium phosphate is hydrated (with 12 water molecules), we use the molar mass of Na\u2083PO\u2084\u00b712H\u2082O in our calculation. Multiplying the moles by this molar mass gives the total mass of sodium phosphate that reacted, which is also the amount that was present in the mixture because it was the limiting reactant.<\/p>\n\n\n\n This process demonstrates how stoichiometry helps relate the mass of reactants and products using mole ratios and molar masses. The final answer, based on careful calculations and proper rounding, is 0.968 grams of sodium phosphate dodecahydrate, corresponding to option (d)<\/strong>.<\/p>\n\n\n\n The Ba3(PO4)2 (molar mass = 602.0 g\/mol) precipitate that formed from a salt mixture has a mass of 0.767 g. If the Na3PO4\u00c2\u00b712H2O (molar mass = 380.2 g\/mol) was the limiting reactant and BaCl2\u00c2\u00b72H2O was the excess reactant in the mixture, what is the mass of Na3FPO4\u00c2\u00b712H2O in the mixture? Select one: a) 0.812 g […]<\/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-31456","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31456","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=31456"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31456\/revisions"}],"predecessor-version":[{"id":31458,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/31456\/revisions\/31458"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=31456"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=31456"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=31456"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Moles of Ba\u2083(PO\u2084)\u2082 formed<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 2: Mole Ratio Between Ba\u2083(PO\u2084)\u2082 and Na\u2083PO\u2084<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 3: Mass of Na\u2083PO\u2084\u00b712H\u2082O in the mixture<\/strong><\/h3>\n\n\n\n
\n\n\n\nCorrect Answer: d) 0.968 g<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation <\/strong><\/h3>\n\n\n\n
![\"\"](\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-341.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"