Thank you for visiting According to the following reaction how many grams of sulfur are formed when 37 4 g of water are formed 2 text H 2 text. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Final answer:
The mass of sulfur formed when 37.4 g of water are formed is 99.8 grams.
Explanation:
To calculate the mass of sulfur formed, we need to use the stoichiometry of the reaction. The stoichiometry tells us the ratio of the coefficients of the reactants and products. In this case, the stoichiometry is 3:2, which means that for every 3 moles of sulfur formed, 2 moles of water are formed.
First, we need to convert the given mass of water to moles. The molar mass of water (H2O) is 18.015 g/mol. Therefore, the number of moles of water formed is:
moles of water = given mass of water / molar mass of water
moles of water = 37.4 g / 18.015 g/mol = 2.073 moles
Now, we can use the stoichiometry to calculate the moles of sulfur formed:
moles of sulfur = (moles of water) * (3 moles of sulfur / 2 moles of water)
moles of sulfur = 2.073 moles * (3 moles of sulfur / 2 moles of water) = 3.1095 moles
Finally, we can convert the moles of sulfur to grams:
grams of sulfur = moles of sulfur * molar mass of sulfur
The molar mass of sulfur (S) is 32.06 g/mol. Therefore, the mass of sulfur formed is:
grams of sulfur = 3.1095 moles * 32.06 g/mol = 99.8 grams
Learn more about calculating the mass of sulfur formed in a chemical reaction here:
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