1 Answers
Call
Talk Doctor Online in Bissoy App
How would you determine the number of moles of $N_2$ that are required to produce 12 mol of $NH_3$ using the equation: $N_2+ 3H_2 -> 2NH_3$?
There is no coefficient in front of the $N_2$ in the reactants. This means that there is one mole of $N_2$ required in the balanced equation. There is a...
1 Answers 1 views
Given the equation:$N_2+3H_2 rarr 2NH_3$, what volume of $NH_3$ at STP is produced if 25 grams of $N_2$ is reacted with an excess of $H_2$?
One thing I think it's interesting to note is that you'll never be able to have that reaction happen in the standard temperature and pressure; you can, however, have the...
1 Answers 1 views
How many grams of $N_2$ are required to completely react with 3.03 grams of $H_2$ for the equation $N_2 + 3H_2 -> 2NH_3$?
Notice the coefficients of $N_2$ and $H_2$. Since the equation has an $N_2$ and $3H_2$, $3$ times as many moles of $H_2$ will be used as $N_2$ assuming a complete...
1 Answers 1 views
Nitrogen (N) and hydrogen (H) react wth each other to produce ammonia ($NH_3$). This reactor is shown in an equation as $N_2 + 3H_2 -> 2NH_3$. What is the ratio of nitrogen atoms to hydrogen atoms in ammonia?
The chemical formula for ammonia is $2NH_3$ meaning that we have two nitrogen atoms and six hydrogen atoms. To find the ratio of nitrogen atoms to hydrogen atoms...
1 Answers 1 views
In the equation $N_2(g) + 3H_2(g) -> 2NH_3(g)$, 8.5 grams of hydrogen and 15.8 grams of nitrogen react to form ammonia. How many grams of ammonia are produced?
Moles of $N_2$ reactant $=$ $(15.8*g)/(28.02*g*mol^-1)$ $=$ $0.564$ $mol$. Moles of $H_2$ reactant $=$ $(8.5*g)/(2.02*g*mol^-1)$ $=$ $4.21$ $mol$. Dihydrogen is the reagent in excess (clearly!). Given that dinitrogen is the...
1 Answers 1 views
How many grams of ammonia, $NH_3$, would be formed from the complete reaction of 4.50 moles hydrogen, $H_2$ in the reaction $N_2 + 3H_2 -> 2NH_3$?
You have almost solved this problem yourself, because you have written a stoichiometrically balanced equation, which establishes equivalent mass: $N_2(g) + 3H_2(g)rightleftharpoons 2NH_3(g)$ This tells us that $28*g$ of dinitrogen...
1 Answers 1 views
In the reaction $N_2(g) + 3H_2(g) -> 2NH_3(g)$, $N_2 = 10.00 M$, $H_2 = 8.00 M$, and $NH_3 = 2.00 M$. What is the equilibrium constant $K$, for this reaction?
Before doing any calculation, take a look at the values of the equilibrium concentrations for the three chemical species that take part in the reaction. Notice that the concentrations of...
1 Answers 1 views
What is $K_(eq)$ for the reaction $N_2+3H_2 -> 2NH_3$ if the equilibrium concentrations are $[NH_3] = 3.0 M, [N_2] = 2.0M$, and $[H_2] = 1.0M$?
As you know, the equilibrium constant for a given chemical equilibrium depends on the equilibrium concentrations of the chemical species that take part in the reaction the stoichiometric...
1 Answers 1 views
What is $K_(eq)$ for the reaction $N_2 + 3H_2 rightleftharpoons 2NH_3$ if the equilibrium concentrations are $[NH_3] = 3 M$, $[N_2] = 2 M$, and $[H_2] = 1 M$?
Use the definition of $K_(eq)$. $K_(eq) = (["products"]^(d,e,f, . . . ))/(["reactants"]^(a,b,c,. . . ))$ $= (["NH"_3]_(eq)^2)/(["H"_2]_(eq)^3["N"_2]_(eq))$ Just make sure you remember to use stoichiometric coefficients correctly. $3H_2$...
1 Answers 1 views
At a certain temperature, 4.0 mol $NH_3$ is introduced into a 2.0 L container, and the $NH_3$ partially dissociates to $2NH_3(g)\rightleftharpoonsN_2(g)+3H_2(g)$.
At equilibrium, 2.0 mol $NH_3$ remains. What is the value of $K_c$?
Remember to include the coefficients in the change in concentratio, as well as the exponents. I get $K_c = 1.69$. If $K_p = K_c(RT)^(Deltan_"gas")$, what would $K_p$ be if this...
1 Answers 1 views