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uysha [10]
4 years ago
13

What is the percent by weight of sodium in sodium sulfate (Na2SO4)? (The molar mass of Na = 22.99, S = 32.07, and O = 16.00.) 16

.18% 22.58% 32.37% 45.05% 51.17%
Chemistry
2 answers:
34kurt4 years ago
8 0
The total mass would be 142.05 g/mol. Since sodium is 22.99 g/mol and there are 2 sodium atoms, it would be 45.98 g/mol. Divide 45.98 by 142.05 and you get 32.37%
Snowcat [4.5K]4 years ago
8 0

<u>Answer:</u> The percentage by weight of sodium in sodium sulfate is 32.37 %.

<u>Explanation:</u>

In Na_2SO_4, there are 2 sodium atoms, 1 sulfur atom and 4 oxygen atoms.

To calculate the mass percent of element in a given compound, we use the formula:

\text{Mass percent of sodium}=\frac{\text{Mass of sodium}}{\text{Molar mass of sodium sulfate}}\times 100

Mass of sodium = (2\times 22.99)=45.98g

Mass of sodium sulfate = [(2\times 22.99)+32.07+(4\times 16)]=142.05g

Putting values in above equation, we get:

\text{Mass percent of sodium}=\frac{45.98g}{142.05}\times 100=32.37\%

Hence, the mass percent of sodium in sodium sulfate is 32.37%.

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Explanation:

<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

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For a complete reaction of the 0.084 moles of HBr you need:

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As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

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Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

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And percent yield (Produced moles / Theoretical moles * 100) is:

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