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%.

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

$M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g$

Theoretical yield of water : T

Moles of tungsten(VI) oxide = $\frac{51.5 g}{232 g/mol}=0.2220 mol$

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

$\frac{3}{1}\times 0.2220 mol=0.6660 mol$

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:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%$

The percent yield of a reaction is 48.05%.

Answer 2

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

Explanation: