## Chemistry and Chemical Reactivity (9th Edition)

a) The number of moles of $ClF_3$ is: $n=PV/RT$ $n=(250/760×1\ atm)×2.5\ L/((0.082)(293\ K))$ $n=0.034\ mol$ From stoichiometry, we find: $0.034×6/4=0.051\ mol\ NiO$ $0.051\ mol×74.69\ g/mol=3.83\ g$ b) Since T and V are constant, it follows: $P=250\ mmHg×(2+3)/4=312.5\ mmHg$ $P_{Cl_2}=250\ mmHg×2/4=125\ mmHg$ $P_{O_2}=250\ mmHg×3/4=187.5\ mmHg$