95.80 g of LiOH are completely reacted with an excess of the other reactant.
If the percent yield is found to be 92.88 %, how many grams of H2O are actually produced?
2007-04-13 05:23:12 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Molar mass of LiOH = 7+16+1 = 24 g
Thus mass of water expected = 18*95.8/24 = 71.85 g
Actual water formed = .9288x71.85 = 66.73 g
2007-04-13 19:45:46 · answer #1 · answered by ag_iitkgp 7 · 0⤊ 0⤋
stability the equation to discover the mol ratio 2LiOH + CO2 ==> Li2CO3 + H2O now you have a a million:a million mol ratio so 88.02 g of CO2 (12.01 + sixteen + sixteen=40 4.01g) is two mol so then discover the form of grams interior the molar mass of Li2CO3 6.ninety 4+6.ninety 4+12.01+sixteen+sixteen+sixteen=seventy 3.89g Now as a results of fact you mol ratio is a million:a million and you have 2 moles of CO2 you need to multiply the Li2CO3 molar mass with the aid of 2 as properly. seventy 3.89g x 2 = 147.78g for this reason there are 147.78g of Li2CO3 produced interior the reaction
2016-12-16 04:46:47 · answer #2 · answered by ? 4 · 0⤊ 0⤋