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OpenStudy (anonymous):
How many grams of nickel tretracarbonyl are needed to produce 1350L of CO at 3000mmHG and 50celsius? Having trouble finding the formula for this problem.
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OpenStudy (anonymous):
pV=nRT -> n(CO) = pV/RT (p in Pa, V in m3, T in K); n(Ni(CO)4)=1/4 n(CO); m(Ni(CO)4)= n(Ni(CO)4)*M(Ni(CO)4) = 8582,17 g
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