Calculate the mass and volume of zinc required to give rise to 50 grams of zinc sulphate when zinc reacts with sulphuric acid.(at.wt of zn-65u,at.wt of S 32u

Answer 1

This question pre-supposes one has knowledge of the reaction between sulfuric acid and Zinc.

Which might be more likely study of chemistry rather than physics, although physics underlies the chemistry as we are talking about the interaction of electrons in bonds and electrodynamics on a macroscopic scale.

First we need to determine the molecular formula for a molecule of zinc sulfate. A quick check of internet materials shows that this is ZnSO_4. Next, what is the percent by weight of zinc in a mole of zinc sulfate.

We see that the periodic table reports the expected weights for a mole of elemental Zn to be 65.39 grams, a mole of elemental oxygen to be 16.00 grams, and a mole of elemental sulfur to be 32.07 grams. Then the percent by mass of Zn in ZnSO₄ is [ 65.39 g / (65.39 +32.07 +4x(16.00)) ] x 100 which is 40.5 percent.

I am given a quantity of zinc sulfate to really only one sig figure as we are not shown that the mass is 50. grams but just 5x10¹ grams, but lets assume the question is giving a mass of 5.0 x10¹ grams. Then this number calculated for the percent by mass 40.5 % will enable a useful prediction for the mass of zinc required to produce the mass of zinc sulfate of 50. grams.

We now have the percent by weight of zinc in zinc sulfate being 40.5 percent and the total mass of our zinc sulfate being 50 g. Mass of zinc in 50 grams of zinc sulfate is 50 g x(40.5/100) = 20 grams.

20 grams of zinc is required with excess sulfuric acid to obtain a theoretical yield of 50 grams of zinc sulfate.

Finally, we are at the point where we need to understand that in making zinc sulfate from zinc and sulfuric acid one zinc atom is involved as a reactant and this produces one zinc sulfate molecule as the product.. This is the chemistry....such that one mole of zinc sulfate requires one mole of Zn and excess sulfuric acid.

Alternatively,

Here we are told that we have 50 grams of zinc sulfate as the product. We need to convert this to moles of Zinc sulfate. We have already calculated the Molecular weight of a mole of zinc sulfate...that is the denominator in our weight percent expression 161.46 grams. So our question is really what percent of a mole or fraction of a mole is 50 grams of zinc sulfate. All one need do is find the ratio between 50 grams and 161.46 grams or (50/161.46) = 0.31 or we have theoretically 0.31 moles of zinc sulfate as our product. This required 0.31 moles of elemental zinc with a scaled mass of 0.31 x 65.39 grams = 20 grams Zn and excess acid.

And one should always reread the question before submitting an answer...

The volume of zinc required requires using the mixed scale factor of density for the metal.

The density of zinc listed  is 7.140 g/mL,. Meaning that 7.140 grams of zinc displaces 1 mL of water.

20 grams/7.140 grams scales this by a factor of 2,8 so that 20 grams displaces 2,8 mL of water and the volume of zinc metal required for this reaction is 2.8 cubic centimeters.