# 1 mole nitrogen gas and 3 moles hydrogen gas form .05 moles NH3 @ equilibrium. What is Kc of the reaction?

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In a sealed 1 L container. I have (NH3)^3/((N2)(H2)^3).

The answer says:

1 mole N2 and 3 moles H2, some amount x of N2 will react with 3x amount of H2 to form 2x amount of ammonia, due to stoichiometric coefficients.

The amount of ammonia at equilibrium is 0.05 M, so 2x 0.05, and x = 0.025.

*Thus [N2] at equilibrium is 1-0.025 = 0.975 M, and [H2] is 3 - 3 x 0.025 M.

Why is anything subtracted from the stoichiometric coefficients at " * "?

Thanks

In a sealed 1 L container. I have (NH3)^3/((N2)(H2)^3).

The answer says:

1 mole N2 and 3 moles H2, some amount x of N2 will react with 3x amount of H2 to form 2x amount of ammonia, due to stoichiometric coefficients.

The amount of ammonia at equilibrium is 0.05 M, so 2x 0.05, and x = 0.025.

*Thus [N2] at equilibrium is 1-0.025 = 0.975 M, and [H2] is 3 - 3 x 0.025 M.

Why is anything subtracted from the stoichiometric coefficients at " * "?

Thanks

This is why we subtract x from the reactants and add x to the products (in their specified ratios from the stoichiometric coefficients). Essentially with our ICE box, the equilibrium is equal to the initial plus the change (making sure we have the right change for the reactants and products since they will be opposite in sign).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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