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There is no coefficient in front of the
There is a 2 coefficient in front of the
This is a 1:2 or 2:1 ratio. Since the desired outcome is
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Using the following reaction, if 1.75 moles of nitrogen gas are reacted with excess hydrogen gas, how many moles of $NH_3$ will be produced? $N_2 + 3H_2 -> 2NH_3$?
Simply write out ratio of the known and unknown components: $(n(NH_3))/(n(N_2))=2/1$ $n(NH_3)=2n(N_2)=2xx1.75mol=3.5mol$
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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...
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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...
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For the reaction represented by the equation $N_2 + 3H_2 -> 2NH_3$, how many moles of nitrogen are required to produce 18 mol of ammonia?
The reaction is: $N_2(g) + 3H_2(g)->2HN_3(g)$ The molar ratio between Nitrogen and ammonia is $1:3$, therefore, to produce 18 moles of ammonia, we will need: $?molN_2=18cancel(molNH_3)xx(1molN_2)/(2cancel(molNH_3))=9molN_2$
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You began with 22 grams of nitrogen, and produced 19 grams of ammonia, $NH_3$, According to the the mass calculations, you should be able to produce 26 grams of ammonia. What is the percent yield? $N_2 + 3H_2 ->2NH_3$
The is 73 %. The formula for percent yield is $color(blue)(bar(ul(|color(white)(a/a) "Percent yield" = "actual yield"/"theoretical yield" × 100 %color(white)(a/a)|)))" "$ ∴ $"Percent yield" = (19 color(red)(cancel(color(black)("g"))))/(26 color(red)(cancel(color(black)("g")))) ×...
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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...
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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...
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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...
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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. $3H_2$...
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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...
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