yacht: limited reactant problem.

[MikeMike]

if 1.65 g of Cu(NO3)2 are obtained from allowing .93 g of Cu to react w/ excess HNO3 what is the percent yield?

i am completely stumped. i know i need to obtain mole ratios, andi know how to do percent yield. however i cant figure out everything exactly

lemme say .93 g of Cu is .0146 moles. thus excess NO3 is atleast .0292.

do i then figure out how many grams of NO3 .0292 is? and then subtract that from Cu(NO3)2 that is, and thus get how many moles of CU there is and then subtract again and get - moles of Cu and then figure out percent yield from there?

or anyhelp?

MIKE

 

[minendo]

Find the limiting reagent.

Calculate molar masses of Cu, Cu(NO3)2, and HNO3. Then use the given numbers (converted to molar mass) to calculate the theoretical and actual yield.

If I had a periodic table or the mass of Cu I could help.

 

[MikeMike]

cu = 63.546 n = 14.00674, o=15.9994, h= 1.00794

MIKE