Question

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

Answer 1

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

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of water = 4.05 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

`\text{Moles of water}=(4.05g)/(18g/mol)=0.225mol`

For the given chemical reaction:

`2H_2O\rightarrow 2H_2+O_2`

• For hydrogen:

By Stoichiometry of the reaction:

2 moles of water is producing 2 moles of hydrogen gas

So, 0.225 moles of water will produce =
`(2)/(2)* 0.225=0.225mol` 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:

`0.225mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=0.45g`

• For nitrogen:

By Stoichiometry of the reaction:

2 moles of water is producing 1 mole of nitrogen gas

So, 0.225 moles of water will produce =
`(1)/(2)* 0.225=0.1125mol` 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:

`0.1125mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=3.15g`

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