first and second reactants.?

For each of the following reactions, calculate the grams of indicated product when 17.5 g of the first reactant and 10.3 g of the second reactant is used:

Part A

4Li(s)+O2(g)→2Li2O(s) (Li2O)

m(Li2O) = 

2 Answers

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  • david
    Lv 7
    2 months ago

    O2 is Lim. reactant

       10.3/32  X 2  X  13.8 g  =  8.884 g  <<< round for sig figs

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  • Fern
    Lv 7
    2 months ago

    17.5 g Li x 1mole Li/6.94 g x 2 moles Li2O/4 moles Li = 1.26 moles Li2O

    10.3 g O2 x 1mole O2/32.0 g x 2 moles Li2O/1 mole O2 = 0.644 moles Li2O

    Limiting reactant is O2:

    0.644 moles Li2O x 29.9 g Li2O/1mole Li2O = 19.3 grams Li2O

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