For the following reaction, calculate how many grams of each product are formed when 4.05 g of water is used.
2 h20 > 2 h2 + o2

The mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.

Explanation:

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

     .....(1)

Given mass of water = 4.05 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

For the given chemical reaction:

By Stoichiometry of the reaction:

2 moles of water is producing 2 moles of hydrogen gas

So, 0.225 moles of water will produce = of hydrogen gas.

Now, calculating the mass of hydrogen gas by using equation 1, we get:

Moles of hydrogen gas = 0.225 mol

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

By Stoichiometry of the reaction:

2 moles of water is producing 1 mole of nitrogen gas

So, 0.225 moles of water will produce = of nitrogen gas.

Now, calculating the mass of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.1125 mol

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

Hence, the mass of hydrogen gas and nitrogen gas produced is 0.45 g and 3.15 g respectively.