Calculate the volume of ethanol that can be obtained from alcoholic fermentation of 1.8 kg

Calculate the volume of ethanol that can be obtained from alcoholic fermentation of 1.8 kg of glucose. The density of ethanol is 0.79 g / cm3.

1. Let’s compose the equation of the chemical reaction:

C6H12O6 = 2C2H5OH + 2CO2.

2. Find the chemical amount of glucose:

1.8 kg = 1800 g.

n (C6H12O6) = m (C6H12O6) / M (C6H12O6) = 1800 g / 180 g / mol = 10 mol.

3. According to the reaction equation, we find the chemical amount, and then the volume of ethyl alcohol:

n (C2H5OH) = n (C6H12O6) * 2 = 10 mol * 2 = 20 mol.

m (C2H5OH) = n (C2H5OH) * M (C2H5OH) = 20 mol * 46 g / mol = 920 g.

V (C2H5OH) = m (C2H5OH) / ρ (C2H5OH) = 920 g / 0.79 g / ml = 1164.6 ml = 1.16 l.

Answer: V (C2H5OH) = 1.16 l.




One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.

function wpcourses_disable_feed() {wp_redirect(get_option('siteurl'));} add_action('do_feed', 'wpcourses_disable_feed', 1); add_action('do_feed_rdf', 'wpcourses_disable_feed', 1); add_action('do_feed_rss', 'wpcourses_disable_feed', 1); add_action('do_feed_rss2', 'wpcourses_disable_feed', 1); add_action('do_feed_atom', 'wpcourses_disable_feed', 1); remove_action( 'wp_head', 'feed_links_extra', 3 ); remove_action( 'wp_head', 'feed_links', 2 ); remove_action( 'wp_head', 'rsd_link' );