How many moles of Na2SO3 can be produced from 32.07 g of sulfur<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n Correct Answer: 1 mole of Na\u2082SO\u2083<\/strong><\/p>\n\n\n\n Explanation:<\/strong><\/p>\n\n\n\n To determine how many moles of sodium sulfite (Na\u2082SO\u2083) can be produced from 32.07 grams of sulfur, begin by examining the chemical relationship and perform a stoichiometric calculation.<\/p>\n\n\n\n Sulfur (S) has an atomic mass of approximately 32.07 g\/mol. Sodium sulfite (Na\u2082SO\u2083) consists of two sodium (Na) atoms, one sulfur (S) atom, and three oxygen (O) atoms. The key part of the question is identifying how many moles of Na\u2082SO\u2083 can be formed from 1 mole of sulfur.<\/p>\n\n\n\n From the chemical formula Na\u2082SO\u2083, it is clear that each molecule of sodium sulfite contains one atom of sulfur. Therefore, 1 mole of sulfur can produce exactly 1 mole of Na\u2082SO\u2083, assuming there is an excess of sodium and oxygen available for the reaction.<\/p>\n\n\n\n Thus, 1 mole<\/strong> of sodium sulfite (Na\u2082SO\u2083) can be produced from 32.07 grams<\/strong> of sulfur.<\/p>\n\n\n\n Stoichiometry involves the calculation of reactants and products in chemical reactions. This problem assumes ideal conditions, such as complete reaction and no limiting reagents besides sulfur. The limiting reagent concept is essential in stoichiometry because it identifies the substance that will run out first, stopping the reaction. In this case, sulfur is the limiting reagent, and the assumption is made that enough sodium and oxygen are available to complete the reaction to form Na\u2082SO\u2083.<\/p>\n","protected":false},"excerpt":{"rendered":" How many moles of Na2SO3 can be produced from 32.07 g of sulfur The correct answer and explanation is: Correct Answer: 1 mole of Na\u2082SO\u2083 Explanation: To determine how many moles of sodium sulfite (Na\u2082SO\u2083) can be produced from 32.07 grams of sulfur, begin by examining the chemical relationship and perform a stoichiometric calculation. Step […]<\/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-34680","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/34680","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=34680"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/34680\/revisions"}],"predecessor-version":[{"id":34684,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/34680\/revisions\/34684"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=34680"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=34680"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=34680"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Determine the molar mass of sulfur<\/h3>\n\n\n\n
So, Moles of sulfur=32.07 g32.07 g\/mol=1.00 mol\\text{Moles of sulfur} = \\frac{32.07\\ \\text{g}}{32.07\\ \\text{g\/mol}} = 1.00\\ \\text{mol}<\/p>\n\n\n\nStep 2: Understand the composition of sodium sulfite<\/h3>\n\n\n\n
Step 3: Final answer<\/h3>\n\n\n\n
Additional Notes:<\/h3>\n\n\n\n