Answer:
The percent yield of a reaction is 48.05%.
Explanation:
![WO_3+3H_2\rightarrow W+3H_2O](https://tex.z-dn.net/?f=WO_3%2B3H_2%5Crightarrow%20W%2B3H_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\times V = 1.00 g/mL\times 5.76 mL=5.76 g](https://tex.z-dn.net/?f=M%3Dd%5Ctimes%20V%20%3D%201.00%20g%2FmL%5Ctimes%205.76%20mL%3D5.76%20g)
Theoretical yield of water : T
Moles of tungsten(VI) oxide = ![\frac{51.5 g}{232 g/mol}=0.2220 mol](https://tex.z-dn.net/?f=%5Cfrac%7B51.5%20g%7D%7B232%20g%2Fmol%7D%3D0.2220%20mol)
According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:
![\frac{3}{1}\times 0.2220 mol=0.6660 mol](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B1%7D%5Ctimes%200.2220%20mol%3D0.6660%20mol)
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}}\times 100](https://tex.z-dn.net/?f=%5C%25%5Ctext%7B%20yield%7D%3D%5Cfrac%7B%5Ctext%7BExperimental%20yield%7D%7D%7B%5Ctext%7BTheoretical%20yield%7D%7D%5Ctimes%20100)
![=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bm%7D%7BT%7D%5Ctimes%20100%3D%5Cfrac%7B5.76%20g%7D%7B11.988%20g%7D%5Ctimes%20100%3D48.05%5C%25)
The percent yield of a reaction is 48.05%.