Compute the mass of caso4 that can be prepared by the reaction of 3.2900g of h2so4 with 3.1660g of caco3

4.3020 grams of calcium sulfate will be produced.

Explanation:

To calculate the number of moles, we use the formula:

   ....(1)

Given mass of calcium carbonate = 3.1660 g

Molar mass of calcium carbonate = 100.0869 g/mol

Putting values in above equation, we get:

Given mass of sulfuric acid = 3.2900 g

Molar mass of sulfuric acid = 98.079 g/mol

Putting values in above equation, we get:

For the given reaction:

By Stoichiometry of the reaction:

1 mole of calcium carbonate reacts with 1 mole of sulfuric acid

So, 0.0316 moles of calcium carbonate will react with =

As, the given amount of sulfuric acid is more than the required amount. Hence, it is present in excess and is considered as an excess reagent.

Calcium carbonate is considered as a limiting reagent because it limits the formation of product.

1 mole of calcium carbonate produces 1 mole of calcium sulfate

So, 0.0316 moles of calcium carbonate will produce = of calcium sulfate.

Now, to calculate the mass of calcium sulfate, we use equation 1:

Molar mass of calcium sulfate = 136.14 g/mol

Moles of calcium sulfate = 0.0316mol

Putting values in equation 1, we get:

Hence, 4.3020 grams of calcium sulfate will be produced.

i'm assuming that the correct answer is indeed the first one/ "a"..

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.