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koban [17]
2 years ago
12

How many grams of O2 are reacted when 3.5 mols of P203 are formed?

Chemistry
1 answer:
Neporo4naja [7]2 years ago
7 0
1.45 moles of P2O3 contains 1.45 moles of O3 which means 4.35 moles of O:

1.45 × 3 = 4.35 moles of O

Then you can calculate moles of O2 by dividing it by 2:

4.35 moles ÷ 2 = 2.175 moles of O2 needed
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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 m
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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%.

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