Solitaire24
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The percentage yield for the following reaction is 70%. calculate the mass of Mg needed to make "125g" of "Mg3N2"
3Mg+N2 - Mg3N2
Last edited by Solitaire24; 4 weeks ago
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STARFIRE13245
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so do you need help or is this your answer can you be more accurate please to help you
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Solitaire24
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(Original post by STARFIRE13245)
so do you need help or is this your answer can you be more accurate please to help you
I would like to know the answer, meaning i need help with the question
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STARFIRE13245
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ok i will help you
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Solitaire24
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(Original post by STARFIRE13245)
ok i will help you
Okay, Thank you (:
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STARFIRE13245
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no problem please wait a minute until i help you ok by the way are you a girl or boy
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GabiAbi84
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What have you done with it so far?
Where are you stuck?
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Solitaire24
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(Original post by STARFIRE13245)
no problem please wait a minute until i help you ok by the way are you a girl or boy
A girl
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Solitaire24
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(Original post by GabiAbi84)
What have you done with it so far?
Where are you stuck?
Actually I think the answer is 63g but, I'm not quite sure
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5hyl33n
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(Original post by STARFIRE13245)
no problem please wait a minute until i help you ok by the way are you a girl or boy
What does that matter? :confused:
Last edited by 5hyl33n; 4 weeks ago
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STARFIRE13245
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just asking do not get me wrong
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STARFIRE13245
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same its 63 not prety suere of my answer
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Solitaire24
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(Original post by STARFIRE13245)
same its 63 not prety suere of my answer
Thank you for trying it out
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STARFIRE13245
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did it work or not?
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STARFIRE13245
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i wish this is helpful
The equation for percent yield is:

percent yield = (actual yield/theoretical yield) x 100%

Where:

actual yield is the amount of product obtained from a chemical reaction
theoretical yield is the amount of product obtained from the stoichiometric or balanced equation, using the limiting reactant to determine product
Units for both actual and theoretical yield need to be the same (moles or grams).

For example, the decomposition of magnesium carbonate forms 15 grams of magnesium oxide in an experiment. The theoretical yield is known to be 19 grams. What is the percent yield of magnesium oxide?

MgCO3 → MgO + CO2

The calculation is simple if you know the actual and theoretical yields. All you need to do is plug the values into the formula:

percent yield = actual yield / theoretical yield x 100%

percent yield = 15 g / 19 g x 100%

percent yield = 79%

Usually, you have to calculate the theoretical yield based on the balanced equation. In this equation, the reactant and the product have a 1:1 mole ratio, so if you know the amount of reactant, you know the theoretical yield is the same value in moles (not grams!). You take the number of grams of reactant you have, convert it to moles, and then use this number of moles to find out how many grams of product to expect.
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Solitaire24
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(Original post by STARFIRE13245)
i wish this is helpful
The equation for percent yield is:

percent yield = (actual yield/theoretical yield) x 100%

Where:

actual yield is the amount of product obtained from a chemical reaction
theoretical yield is the amount of product obtained from the stoichiometric or balanced equation, using the limiting reactant to determine product
Units for both actual and theoretical yield need to be the same (moles or grams).

For example, the decomposition of magnesium carbonate forms 15 grams of magnesium oxide in an experiment. The theoretical yield is known to be 19 grams. What is the percent yield of magnesium oxide?

MgCO3 → MgO + CO2

The calculation is simple if you know the actual and theoretical yields. All you need to do is plug the values into the formula:

percent yield = actual yield / theoretical yield x 100%

percent yield = 15 g / 19 g x 100%

percent yield = 79%

Usually, you have to calculate the theoretical yield based on the balanced equation. In this equation, the reactant and the product have a 1:1 mole ratio, so if you know the amount of reactant, you know the theoretical yield is the same value in moles (not grams!). You take the number of grams of reactant you have, convert it to moles, and then use this number of moles to find out how many grams of product to expect.
It did work and I cannot say how much of a help you were so, thanks a ton
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STARFIRE13245
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your welcome any time.i am here to help anytime
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