To calculate the moles of hydrogen gas produced, we need to begin with the balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid (HCl):<\/p>\n\n\n\n
Balanced equation:<\/strong><\/p>\n\n\n\n 2Al+6HCl\u21922AlCl3+3H22Al + 6HCl \\rightarrow 2AlCl_3 + 3H_22Al+6HCl\u21922AlCl3\u200b+3H2\u200b<\/p>\n\n\n\n This shows that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas.<\/p>\n\n\n\n 1. Find the molar mass of aluminum (Al):<\/strong> 2. Calculate moles of aluminum used:<\/strong>Moles of Al=1.063 g26.98 g\/mol=0.0394 mol (rounded to 4 decimal places)\\text{Moles of Al} = \\frac{1.063\\ \\text{g}}{26.98\\ \\text{g\/mol}} = 0.0394\\ \\text{mol (rounded to 4 decimal places)}Moles of Al=26.98 g\/mol1.063 g\u200b=0.0394 mol (rounded to 4 decimal places)<\/p>\n\n\n\n 3. Use the mole ratio from the balanced equation to find moles of H\u2082:<\/strong> So the mole ratio is:3 mol H22 mol Al\\frac{3\\ \\text{mol H}_2}{2\\ \\text{mol Al}}2 mol Al3 mol H2\u200b\u200b<\/p>\n\n\n\n Now use this ratio to calculate moles of hydrogen:Moles of H2=0.0394 mol Al\u00d732=0.0591 mol H2\\text{Moles of H}_2 = 0.0394\\ \\text{mol Al} \\times \\frac{3}{2} = 0.0591\\ \\text{mol H}_2Moles of H2\u200b=0.0394 mol Al\u00d723\u200b=0.0591 mol H2\u200b<\/p>\n\n\n\n Moles of H\u2082 Gas = 0.0591 mol<\/strong><\/p>\n\n\n\n <\/p>\n\n\n\n The reaction between aluminum and hydrochloric acid is a single displacement reaction where aluminum displaces hydrogen from the acid. The balanced chemical equation provides the molar relationship between reactants and products. Here, two moles of aluminum yield three moles of hydrogen gas. This ratio is essential for solving the problem accurately.<\/p>\n\n\n\n We start by converting the given mass of aluminum into moles using its molar mass. The molar mass of aluminum is approximately 26.98 grams per mole. By dividing the given mass of 1.063 grams by this molar mass, we find how many moles of aluminum were used in the reaction. This results in about 0.0394 moles of aluminum.<\/p>\n\n\n\n Next, using the stoichiometry from the balanced equation, we determine how much hydrogen gas is formed. For every two moles of aluminum, three moles of hydrogen gas are produced. This means the ratio of hydrogen gas to aluminum is three to two. We multiply the moles of aluminum by this ratio to find the number of moles of hydrogen gas generated.<\/p>\n\n\n\n Thus, multiplying 0.0394 moles of aluminum by 1.5 (which is three divided by two) gives us approximately 0.0591 moles of hydrogen gas. This is the amount of hydrogen gas that would be produced under ideal conditions if all the aluminum reacted completely with excess hydrochloric acid.<\/p>\n\n\n\n This type of stoichiometry calculation is commonly used in chemistry to predict the amounts of products formed in chemical reactions based on reactant quantities.<\/p>\n\n\n\n Calculate the moles of hydrogen gas produced by the reaction of 1.063 g of aluminum with hydrochloric acid. Moles of H2 Gas = The Correct Answer and Explanation is: To calculate the moles of hydrogen gas produced, we need to begin with the balanced chemical equation for the reaction between aluminum (Al) and hydrochloric acid […]<\/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-36231","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36231","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=36231"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36231\/revisions"}],"predecessor-version":[{"id":36233,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36231\/revisions\/36233"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36231"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36231"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36231"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep-by-step Calculation:<\/h3>\n\n\n\n
Al = 26.98 g\/mol<\/p>\n\n\n\n
From the balanced equation:2 mol Al\u21923 mol H22\\ \\text{mol Al} \\rightarrow 3\\ \\text{mol H}_22 mol Al\u21923 mol H2\u200b<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\n\n\n\nExplanation <\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"