**Balanced equation**

#"K"_2"Cr"_2"O"_7 + "K"_2"CO"_3"##rarr##"2K"_2"CrO"_4 + "CO"_2"#

Determine mol #"K"_2"CO"_3"# are in #"1.00 g"#. (#"1.00 g"# is arbitrary, since the number of significant figures were not specified. You can redo the calculations with the number of significant figures you wish.) Divide the mass of #"K"_2"CO"_3"# by its molar mass #("138.205 g/mol")# by multiplying the mass of #"K"_2"CO"_3"# by the inverse of its molar mass #("1 mol"/"138.205 g")#.

#1.00color(red)cancel(color(black)("g K"_2"CO"_3))xx(1"mol K"_2"CO"_3)/(138.205color(red)cancel(color(black)("g K"_2"CO"_3)))="0.007236 mol K"_2"CO"_3"#

I am keeping an extra digit to reduce rounding errors. I will round to three significant figures at the end.

Determine mol #"K"_2"CrO"_4"# by multiplying mol #"K"_2"CO"_3"# by the mol ratio between the two compounds in the balanced equation, with mol #"K"_2"CrO"_4"# in the numerator.

#0.007236color(red)cancel(color(black)("mol K"_2"CO"_3))xx(2"mol K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CO"_3)))="0.01447 mol K"_2"CrO"_4"#

Determine mass #"K"_2"CrO"_4"# by multiplying mol #"K"_2"CrO"_4"# by its molar mass #("194.189 g/mol")#.

#0.01447color(red)cancel(color(black)("mol K"_2"CrO"_4))xx(194.189"g K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CrO"_4)))="2.81 g K"_2"CrO"_4"# (rounded to three significant figures)

For every gram of #"K"_2"CO"_3"# used, #"2.81 g K"_2"CrO"_4"# is formed.