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In the following reaction, 451.4 g of lead reacts with excess oxygen forming 369.6 g of lead(II) oxide. Calculate the percent yield of the reaction.
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The reaction is: \[2Pb(s) + O _{2}(g)-> 2PbO(s)\] I cannor figure out how to do this equation! Please help :(
We can to find the theoretical yield first. Which is like the amount we can get if everything goes perfect. Convert the mass of the lead to moles of lead. Use stoichiometry to find out how much lead (II) oxide you can make from that. Convert the moles of lead II oxide to grams.
Then, find the percent yield: % yield = actual yield/theoretical yield
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