Respuesta :
Answer:
The percent yield of a reaction is 48.05%.
Explanation:
[tex]WO_3+3H_2\rightarrow W+3H_2O[/tex]
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
[tex]M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g[/tex]
Theoretical yield of water : T
Moles of tungsten(VI) oxide = [tex]\frac{51.5 g}{232 g/mol}=0.2220 mol[/tex]
According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:
[tex]\frac{3}{1}\times 0.2220 mol=0.6660 mol[/tex]
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:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
[tex]=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%[/tex]
The percent yield of a reaction is 48.05%.
Answer:The percent yield of a reaction is 48.05%.
Explanation:
