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Alisiya [41]
2 years ago
11

The compound sodium thiosulfate pentahydrate, Na2S2O3 • 5H2O

Chemistry
1 answer:
Mnenie [13.5K]2 years ago
4 0

The theoretical yield is 204.4 g while the percent yield is 2.57%.

<h3>What is theoretical yield?</h3>

Theoretical yield is the amount of product obtained based on the stoichiometry of the reaction.

S8(s) + 8 Na2SO3(aq) + 40 H2O(l) --->8 Na2S2O3·5 H2O(s)

Number of moles of sulfur =  3.25 g /8(32) = 0.013 moles

Number of moles of sodium sulfite =  13.1 g/126 g/mol = 0.103 moles

Since 1 moles of sulfur reacts with 8 moles of sodium sulfite

0.013 moles reacts with 0.013 moles ×  8 moles /1 mole = 0.104 moles

There is not enough sodium sulfite hence it is the limiting reactant.

1 mole of sodium sulfite yields 8 moles of product

0.103 moles of sodium sulfite yields  0.103 moles × 8 moles /1 mole = 0.824 moles

Mass of product = 0.824 moles × 248 g/mol = 204.4 g

percent yield =  5.26 g /204.4 g × 100/1

= 2.57%

Learn more about percent yield: brainly.com/question/2506978

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Name the two products when calcium carbonate (CaCO3) is heated
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If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
3 years ago
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