Question

What is the percent yield of a reaction in which 47.5 g tungsten (VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 9.40 mL water (d = 1.00 g/mL)?

Answer 1

Answer:

The percent yield of a reaction is 85.04%

Explanation:

$:WO_3+3H_2\rightarrow W+3H_2O$

Mass of $WO_3$ = 47.5g

Molar mass of $WO_3$ = 232 g/mole

Molar mass of $H_2O$= 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

$M=d\times V = 1.00 g/mL\times 9.40 mL=9.40 g$

Moles of tungsten(VI) oxide =$\frac{47.5 g}{232 g/mol}=0.2047 mol$

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2047 moles of tungsten(VI) oxide will give:

$\frac{3}{1}\times 0.2047 mol=0.6141 mol$

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 = {$\frac{Experimental yield}{theoretical yield} \times 100$

$\frac{9.40}{11.054} \times 100\\\\= 85.04$

The percent yield of a reaction is 85.04%