Question

If 139 gm of zinc metal and 207 gmof copper II nitrate react, how many gramsof copper metal will be produced for thisreactionZn + Cu(NO3)2 —>Zn(NO3)2 + Cu

Answer 1

`70.47\text{ g}`

Step-by-step explanation:

Here, we want to get the mass of the copper metal produced

From what we have in the balanced equation of reaction, the mole ratio for all is 1 to 1

This is the theoretical mole ratio

To get the actual, we need to look through all

Firstly, let us get the mass produced considering zinc

We need to get the number of moles produced

That would be the mass of zinc divided by the atomic mass of zinc

The atomic mass of zinc is 65 amu

The number of moles is thus: 139/65 = 2.1385 moles

The mass of copper would be the this number of moles multiplied by the atomic mass of copper

The atomic mass of copper is 64 amu

The mass is thus:

`64\text{ }*2.1385\text{ = 136.86 g}`

Now, let us do the same considering copper ii nitrate

We divide the mass by the molar mass

The molar mass of copper ii nitrate is 188 g/mol

The number of moles will be:

`(207)/(188)\text{ = 1.101 mol}`

The mass of copper produced will be the product of the number of moles and the atomic mass of copper. The atomic mass of copper is 64 amu

The mass is thus:

`64*\text{ 1.101 = 70.47 g}`

Copper nitrate produces less amount of copper and thus it is the limiting reagent and its value should be considered