What is Yield Reactions?
In chemistry, yield, also known as reaction yield, is a percentage measure of the number of moles of a product formed in relation to the reactant consumed in a chemical reaction. One of the most important factors that scientists must consider in organic and inorganic chemical synthesis processes is yield. The terms "yield", "conversion" and "selectivity" are used in chemical reaction engineering to describe ratios of how much of a reactant was consumed (conversion), how much desired product was formed (yield), and how much undesired product was formed (selectivity), represented as X, Y, and S.
Main content
Mass of [tex]WO_3=45.5g\\[/tex]
Volume of [tex]H_2O[/tex] produced = 9.60mL
Calculations:
We calculate the actual yield of [tex]H_2O[/tex]. To get the mass of [tex]H_2O[/tex] produced, multiply the given volume of water by its given density.
Mass of [tex]H_2O[/tex] produced (actual) = 9.60mL [tex]H_2O \times \frac{1.00g H_2O}{1mL H_2O } = 9.60g H_2O.\\[/tex]
Next, we translate the statement into the following chemical equation.
[tex]WO_3 (s) + H_2 (g) \rightarrow W(s) + H_2O (l)\\[/tex]
To balance the chemical equation, we place coefficient 3 in front of [tex]H_2[/tex] and [tex]H_2O.\\[/tex]
[tex]WO_3(s)+ 3H_2 (g) \rightarrow W(s) + 3H_2O(l)\\\\[/tex]
To get the theoretical yield of water. we multiply the given mass of [tex]WO_3[/tex] by the reciprocal of its molar mass, the molar ratio between [tex]WO_3[/tex] and [tex]H_2O,[/tex] and the molar mass of water. The molar masses of [tex]WO_3[/tex] and [tex]H_2O[/tex] are 231.84 g/mol and 18.02g/mol. The molar ratio 3 mol [tex]H_2O[/tex] are 231.84 g/mol
and 18.02g/mol. The molar ratio is 3 mol [tex]H_2O/mol WO_3[/tex].
[tex]H_2O[/tex] produced = 45.5g [tex]WO_3 \times \frac{1 mol WO_3}{231.84g WO_3} \times \frac{3 mol H_2O}{1 mol WO_3} \times \frac{18.02g H_2O}{1 mol H_2O}[/tex]
[tex]H_2O[/tex] produced = 10.61g
The precent yield is simply the ratio of the actual yield to the theoretical yield multiplied by 100.
[tex]\% yield = \frac{actual yield}{theoretical yield} \times 100 = \frac{9.60g H_2O}{10.61g H_2O} \times 100 = 90.5\%[/tex]
Therefore the correct answer is 90.5%
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