Question

What is the percent yield of a reaction in which 51.5 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 5.76 mL of water (d = 1.00 g/mL)?

Answer 1

Answer:The percent yield of a reaction is 48.05%.

Step-by-step explanation:

Answer 2

The percent yield of a reaction is 48.05%.

Step-by-step explanation:

`WO_3+3H_2\rightarrow W+3H_2O`

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

`M=d* V = 1.00 g/mL* 5.76 mL=5.76 g`

Theoretical yield of water : T

Moles of tungsten(VI) oxide =
`(51.5 g)/(232 g/mol)=0.2220 mol`

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

`(3)/(1)* 0.2220 mol=0.6660 mol`

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

`=(m)/(T)* 100=(5.76 g)/(11.988 g)* 100=48.05\%`

The percent yield of a reaction is 48.05%.