We can do this by multiplying the volume of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) used (in liters) by its molarity:
\( 36.92 \mathrm{~mL} = 0.03692 \mathrm{~L} \)
\( \mathrm{Moles~of~K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} = 0.03692 \mathrm{~L}
Show more…