Question

Ignoring the CuSO4, and based solely on the volume and molarity of the Na2S(aq), how many grams of CuS(s) would be produced by this

Answer 1

0.019122 grams of CuS(s) would be produced .

Step-by-step explanation:

Given , volume of
`CuSO_(4)` is 4ml = 0.004 L

volume of
`Na_(2)S` is 2ml = 0.002 L

`CuSO_(4) + Na_(2)S $\rightarrow$ Na_(2)SO_(4) + CuS`

Moles of
`CuSO_(4)` = Molarity x volume (L)

= 0.1M x 0.004 L

= 0.0004 moles

Moles of
`Na_(2)S` = 0.1M x 0.002 L

= 0.0002 moles

1 mole of
`Na_(2)S$\rightarrow$` 1 mole of
`CuS`

Therefore , 0.0002 moles
`Na_(2)S $\rightarrow$` 1mole of
`CuS`

Mass of
`CuS` = Moles x molar mass

= 0.0002 x 95.611

= 0.019122 grams .

Hence the answer is 0.019122 grams.