If 25.4 grams of water react to form 2.8 grams of hydrogen, how many grams of oxygen must simultaneously be formed?

There will be formed 22.6 grams O2

Explanation:

Step 1: Data given

Mass of 25.4 grams

Mass of hydrogen formed = 2.8 grams

Step 2: The balanced equation

2H2O(l) → 2H2(aq) + O2(aq)

For 2 moles H2O we'll have 2 moles hydrogen and 1 mol O2

Step 3: Calculate moles of H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 25.4 grmas / 18.02 g/mol

Moles H2O = 1.41 moles

Step 4: Calculate moles hydrogen formed

Moles H2 = 2.8 grams / 2.0 g/mol

Moles H2 =  1.4 moles

Step 5: Calculate moles of O2

For 2 mole of H2O reacted, there will be formed 1 mol of H2O

For 1.41 moles of H2O formed there will be formed 0.705moles O2.

Step 6: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 0.705* 32 g/mol

Mass O2 = 22.6 grams O2

There will be formed 22.6 grams O2

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.