This topic has 1 reply, 1 voice, and was last updated 3 months ago by .Viewing 2 posts - 1 through 2 (of 2 total)AuthorPosts April 23, 2024 at 4:00 pm #3058 Frage: (st-28) ⓘ 🖶 ausdrucken [klick Frage/Antwort] ◀ – ▶ Stelle 51 Gramm Ammoniak (NH3) aus den Elementen her, wie viel Gramm der einzelnen Produkte werden benötigt? Hinweis: Hinweis: Da Ammoniak = NH3, somit wären die Edukte H2 und N2 April 23, 2024 at 4:00 pm #3059 Textantwort: Die ausgeglichene Reaktionsgleichung lautet: 1·N2 + 3·H2 → 2·NH3 Stoff M(g/mol) m(g) n(mol) NH3 17 51 3 H2 2 9 3/2·3 = 4.5 N2 28 42 3/2 = 1.5 Es werden somit 9 Gramm H2 sowie 42 Gramm N2 benötigt.AuthorPostsViewing 2 posts - 1 through 2 (of 2 total)You must be logged in to reply to this topic.Log In Username: Password: Keep me signed in Log In Warning: Trying to access array offset on value of type bool in /home/httpd/vhosts/rainer.ch/www_chemieaufgaben_ch/wp1/wp-content/plugins/bbp-style-pack/includes/functions.php on line 2502 Log in / Register