Suppose 650 mL of hydrogen gas are produced through a displacement reaction involving solid iron and sulfuric acid, H2SO4, at STP. How many grams of iron (II) sulfate are produced? Density H2(g) is on your periodic table. PLEASE SHOW WORK!!

Respuesta :

Answer:

                       4.415 g of FeSO

Explanation:

                    The balance chemical equation for given single replacement reaction is as follow;

                                   Fe + H₂SO₄ → FeSO₄ + H₂

Data Given;

                   Volume =  650 mL  =  0.65 L

                   Density at STP =  0.08988 g/L

                   Mass = Density × Volume = 0.08988 g/L × 0.65 L = 0.0584 g

Step 1: Calculate Moles of H₂ as;  

                   Moles  =  Mass / M.Mass

                   Moles  =  0.0584 g / 2.01 g/mol

                   Moles  =  0.0290 mol of H₂

Step 2: Find out moles of FeSO₄ as;

According to balance chemical equation,

             1 mole of H₂ is produced along with  =  1 mole of FeSO₄

So,

    0.0290 moles of H₂ will be produced along with  =  X moles of FeSO₄

Solving for X.

                      X  =  0.0290 × 1 mol / 1 mol

                     X =  0.0290 moles of FeSO₄

Step 3: Calculate mass of FeSO₄ as;

                    Mass  =  Moles × M.Mass

                    Mass  =  0.0290 mol × 151.90 g/mol

                    Mass  =  4.415 g of FeSO

The mass of iron II sulfate produced is 4.56 g.

What is a reaction?

A reaction is said to occur when substances combine to form a new substance or when a substance is broken down into its components.

The equation of the reaction is; Fe + H2SO4  --->FeSO4  + H2

If 1 mole of gas occupies 22.4 L

x moles of a gas occupies 0.65 L

x = 0.03 moles of H2

1 mole of  Fe(SO4)2 is produced for every 1 moles  of H2

x moles of Fe(SO4)2 is produced for every  0.03 moles of H2

x =0.03 moles

Mass of Fe(SO4)2 = 0.03 moles * 152 g/mol = 4.56 g

Learn more about moles: https://brainly.com/question/4083236