Question
Asked Nov 19, 2019
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What is the percent yield of a reaction in which 63.0 g of tungsten(VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 6.84 mL of water (d = 1.00 g/mL)?
 

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Expert Answer

Step 1

The molar mass of WO3 (M) is 231.84 g/mol. The number of moles in 63.0 g (mass, m) of tungsten oxide (n) is calculated as follows:

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т п 3. М 63.0 g 231.84 g/mol = 0.272 mol

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Step 2

The mass (m’) of 6.84 mL (volume, V) of water is calculated as follows:

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d = V m' 1.0 g/mL 6.84 mL m'=6.84 g

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Step 3

The molar mass of water is 18.02 g/mol. The number of moles (...

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6.84 g n' = 18.02 g/mol = 0.380 mol

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