How many grams of chromium are needed to react with an excess of #CuSO_4# to produce 16.2 g Cu?

1 Answer
Aug 24, 2017

#"29.7 g Cu"#

Refer to the explanation for the process.


Start with a balanced equation.

#"2Cr + 3CuSO"_4"##rarr##"Cr"_2"(SO"_4")"_3 + "3Cu"#

Determine moles Cr from the given mass and the molar mass (51.996 g/mol).

#16.2color(red)cancel(color(black)("g Cr"))xx(1"mol Cr")/(51.996color(red)cancel(color(black)("g Cr")))="0.312 mol Cr"#

Determine moles Cu by multiplying by the mole ratio between Cr and Cu from the balanced equation, with Cu in the numerator.

#0.312color(red)cancel(color(black)("mol Cr"))xx(3"mol Cu")/(2color(red)cancel(color(black)("mol Cr")))="0.468 mol Cu"#

Determine mass Cu by multplying mol Cu by molar mass Cu (63.546 g/mol).

#0.468color(red)cancel(color(black)("mol Cu"))xx(63.546"g Cu")/(1color(red)cancel(color(black)("mol Cu")))="29.7 g Cu"# (rounded to three significant figures)

The steps can be combined into one equation:

#16.2color(red)cancel(color(black)("g Cr"))xx(1color(red)cancel(color(black)("mol Cr")))/(51.996color(red)cancel(color(black)("g Cr")))xx(3color(red)cancel(color(black)("mol Cu")))/(2color(red)cancel(color(black)("mol Cr")))xx(63.546"g Cu")/(1color(red)cancel(color(black)("mol Cu")))="29.7 g Cu"#

Note: This is the mass of copper that can be produced in an excess of copper sulfate in a closed system, and assuming that there are no side reactions and that the reaction went to completion. So If this were an actual experiment that you performed, you would most likely get less copper.