If 1.20 g of Zn is allowed to react with 2.00 g of CuSO4, according to the equation below, how many grams of Zn will remain after the reaction is complete? CuSO4(aq)+Zn(s)→ZnSO4(aq)+Cu(s) 

After the reaction is complete 0.94 g of Zn are still remaining.

Explanation:

We define the reaction as:

CuSO₄ (aq)  +  Zn (s)  →  ZnSO₄ (aq)  +  Cu(s)  

In order to determine the grams of Zn that will remain after the reaction is complete, we need to confirm if the Zn is the limiting reactant.

We determine moles of each reactant:

1.20 g / 65.41 g/mol = 0.0183 moles of Zn

2 g / 159.6 g/mol = 0.0125 moles of sulfate

Ratio is 1:1. 1 mol of sulfate needs 1 mol of Zn to react.

If I have 0.0125 moles of sulfate I will ned the same moles of Zn to complete the reaction.

I have 0.0183 moles, so it is ok that Zn is the reagent in excess

We convert the moles to mass and we make the substraction to answer the grams of Zn that will remain after the reaction:

0.0183 moles . 159.1 g /mol = 2.92 g

0.0125 moles . 159.1 g /mol =  1.98 g

To complete all the reaction we need 2.92 g of sulfate, but I have 1.98 g

After the reaction is complete (2.92 g - 1.98 g) = 0.94 g are still remaining.

it is a solid

explanation: