Answer:
The percent yield of a reaction is 85.04%
Explanation:
[tex]:WO_3+3H_2\rightarrow W+3H_2O[/tex]
Mass of [tex]WO_3[/tex] = 47.5g
Molar mass of [tex]WO_3[/tex] = 232 g/mole
Molar mass of [tex]H_2O[/tex]= 18 g/mole
Volume of water obtained from the reaction , V= 9.40mL
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 9.40 mL=9.40 g[/tex]
Moles of tungsten(VI) oxide =[tex]\frac{47.5 g}{232 g/mol}=0.2047 mol[/tex]
According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2047 moles of tungsten(VI) oxide will give:
[tex]\frac{3}{1}\times 0.2047 mol=0.6141 mol[/tex]
Mass of 0.6141 moles of water:
0.6141 mol × 18 g/mol = 11.054 g
To calculate the percentage yield of reaction , we use the equation:
% yield = {[tex]\frac{Experimental yield}{theoretical yield} \times 100[/tex]
[tex]\frac{9.40}{11.054} \times 100\\\\= 85.04[/tex]
The percent yield of a reaction is 85.04%
