posted by Eli .
A mettallurgist decided to make a special alloy. He melted 30 grams of Au(Gold) with 30 grams of Pb. This melting process resulted to 42 grams of an Au-Pb alloy and some leftover Pb. This scientist wanted to make 420 grams of this alloy with no leftovers of any reactants. how much of Au and Pb would he need to use in order to have the desired grams of the alloy?
Au + Pb = Au-Pb left over
Not a great deal of chemistry here as it is by proportion.
He melted 30 grams of Au (gold) with 30 grams of Pb. This melting process resulted to 42 grams of an Au-Pb alloy and some leftover Pb.
So assume all the gold is incorporated then (42-30) g of lead used, which is 12 g.
Thus for each 42 g of alloy, 30 g of gold and 12 g of lead are needed.
Hence for 420 g (42 g x 10), 300 g (30 g x 10) of gold and 120 g (12 g x 10)of lead are needed.