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)?

The percent yield of a reaction is 85.04%

Explanation:

Mass of = 47.5g

Molar mass of = 232 g/mole

Molar mass of = 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

Moles of tungsten(VI) oxide =

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

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 = {

The percent yield of a reaction is 85.04%

the answer for sodium is 0.186

the answer for rubidium is 0.251

i think it is true since you can test the answer and can use observations to complete it.