Question

Chapter 4 Homework 0 13 2 attempts left Check my work Be sure to answer all parts. Report problem 6.66 points In a combination reaction, 1.58 of lithium is mixed with 6,77 of oxygen. Guided Solution (a) Which reactant is present in excess! Print Reference oxygen (b) How many moles of product are formed? roduct are present (c) After the reaction, how many grams of each reactant an gLi 0; g Lio

Answer 1

mole Solution 4 Li + O2 ->? Ligo de - 4 mole Imole & mole 4 mole x6.949 imele XIag male X 29.887, = 27.76 gli = 32g O2 = 59.16g Lizo @ Molar mass of Li = 6-94g/mole molar mass of Oz = 32g/mole Malar mass of Li₂O = 29.88 g/mole 27.76g Li react with 32g Oz in 1-589 Li react with 32g xl-589 02 - 27.769 = 1.82g Oz Ins- so we can say that Oz (oxygen) will be in excess i.e. 6.779 - 1.82g = 4.95 g Oz is in excess. so we can calculate the product formed either by 1.58g . Li or 1.82g Oz % 27.76 g Li produced 59.76g Lizo is 1.58 g Li produced some produced 59.769 X 1.58g Lizo 27-769 - 3.4 g Lieo oles of lio mass of Lizo melarmass of lico Scanned with CamScanner

moles of Lizo (Broduct) = 30 kglmale = 0.114 male Ans:- 0.114 moles of product (Live) are formed. ♡ Li will not be present after reaction because it will - completely exhausted- Ogli 4.95 g Oz 3.4 g Lizo cs Scanned with CamScanner