Using this equation: 4 Ag + 2 H2S + O2 --> 2 Ag2S + 2 H2O, how many moles of Ag2S will be produced from 6.2 moles of Ag? Please show work
please help me!!
This is a simple stoichiometry problem. I worked a very similar one for Veronica. Here is a link to that work. You already have the equation AND you have grams converted into mols. Follow the other steps.
https://www.jiskha.com/questions/1815058/i-cant-get-this-if-4-00-g-of-h2-are-made-to-react-with-excess-co-g-how-many-grams-of
Post your work if you get stuck.
The balanced chemical equation is: 4 Ag + 2 H2S + O2 → 2 Ag2S + 2 H2O.
From the equation, we can see that 4 moles of Ag will produce 2 moles of Ag2S.
To calculate the moles of Ag2S produced from 6.2 moles of Ag, we can set up a simple ratio:
(6.2 moles of Ag) * (2 moles of Ag2S / 4 moles of Ag)
Simplifying the equation:
(6.2 moles of Ag) * (0.5 moles of Ag2S / 1 mole of Ag)
Calculating the result:
= 6.2 * 0.5
= 3.1 moles of Ag2S
Therefore, 6.2 moles of Ag will produce 3.1 moles of Ag2S.
To answer this question, we need to determine the mole ratio between Ag and Ag2S in the balanced chemical equation. The coefficient of Ag2S in the equation is 2, indicating that for every 4 mol of Ag, 2 mol of Ag2S will be produced.
Using the mole ratio, we can set up a proportion to calculate the number of moles of Ag2S produced from 6.2 moles of Ag.
(4 mol Ag) / (2 mol Ag2S) = (6.2 mol Ag) / (x mol Ag2S)
Cross-multiplying, we get:
4 mol Ag * x mol Ag2S = 2 mol Ag2S * 6.2 mol Ag
Simplifying, we have:
4x mol Ag2S = 12.4 mol Ag2S
Finally, solving for "x" (moles of Ag2S produced), we divide both sides of the equation by 4:
x = 12.4 mol Ag2S / 4
x ≈ 3.1 mol Ag2S
Therefore, approximately 3.1 moles of Ag2S will be produced from 6.2 moles of Ag.