Starting with 1.5052g of bacl2•2h2o and excessh2so4, how many grams of baso4 can be formed?

The mass of barium sulfate produced is 1.45 grams.

Explanation:

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

     .....(1)

Given mass of = 1.5052 g

Molar mass of = 244.26 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of barium chloride and sulfuric acid follows:

As, sulfuric acid is present in excess, it is considered as an excess reagent.

Thus, is considered as a limiting reagent because it limits the formation of product

By Stoichiometry of the reaction:

1 mole of produces 1 mole of barium sulfate

So, 0.0062 moles of will produce = moles of barium sulfate

Now, calculating the mass of barium sulfate by using equation 1:

Molar mass of barium sulfate = 233.4 g/mol

Moles of barium sulfate = 0.0062 moles

Putting values in equation 1, we get:

Hence, the mass of barium sulfate produced is 1.45 grams.

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2