Answer
See the answer below.
Work Step by Step
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:
$P=250\ mmHg×(2+3)/4=312.5\ mmHg$
$P_{Cl_2}=250\ mmHg×2/4=125\ mmHg$
$P_{O_2}=250\ mmHg×3/4=187.5\ mmHg$
