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Given stoichiometric oxygen, ammonia gives an equimolar quantity of
Clearly, dioxygen is in deficiency, and only
And thus only
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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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In this reaction: $4HCl (g) + O_2 (g) -> 2H_2O (l) + 2Cl_2(g)$. When 63.1 g of $HCI$ react with 17.2 g of $O_2$, 49.3 g of $Cl_2$ are collected. How do you determine the limiting reactant, theoretical yield of $Cl_2$ and percent yield for the reaction?
We must first identify the limiting reactant, and then we calculate the theoretical yield and percent yields. We start with the balanced equation. $color(white)(mmmmmmmm)"4HCl" + "O"_2 → "2H"_2"O" +...
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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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Define the terms theoretical yield, actual yield, and percent yield. Why is the actual yield in a reaction almost always less than the theoretical yield? Can a reaction ever have 110% actual yield?
Actual yield in a reaction is almost always less than the theoretical yield, primarily because losses of the substances involved may occur anywhere in an experiment. Otherwise, there can be...
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$2H_2O -> 2H_2 + O_2$ What is the percent yield of $O_2$ if 10.2 g of $O_2$ is produced from the decomposition of 17.0 g of $H_2O$?
You know by looking at the balanced chemical equation $2"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))$ that it takes $2$ moles of water to produce...
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A yield of $NH_3$ of approximately 98% can be obtained at 200 deg C and 1,000 atmospheres of pressure. How many grams of $N_2$ must react to form 1.7 grams of ammonia, $NH_3$?
We always have to read the problems completely to understand the question. In this case even when we have pressure and temperature conditions specified, they don't matter, first because they...
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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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How does ammonia behave as a Bronsted base:
$"(i) "NH_3(g) + 7/2O_2(g) rarr 2NO_2(g) + 3H_2O$
$"(ii) "2NH_3(g) rarr NH_4^+ +NH_2^(-) $
$"(iii) "NH_3(g) + H_3CBr(g) rarr H_3CBr(g) + HBr(g)$
$"(iv) "NH_3(aq) +H_2O(l) rarr NH_4^+ + HO^-$?
Ammonia is a Bronsted base: $NH_3(aq) + H_2O(l) rightleftharpoons NH_4^+ + OH^-$ $pK_b = 4.75$ I am not going to solve this equation using $pK_b$, but most of the ammonia...
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When ammonia, $NH_3$, dissolves in water, hydrogen ions are translerred from the water to ammonia to form ammonium ions and hydroxide ions. Ths reaction can be seen below. Why is ammonia a base in this reaction?
In acting as an electron pair donor, the nitrogen has become quaternized, and has demonstrably increased hydroxide ion concentration. It has therefore acted as a base by several definitions.
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Ammonia ($NH_3$) is a weak base with a $K_b = 1.8 x 10^-5$. What is the balanced chemical equation for the reaction of ammonia with water. Using the I.C.E the method, calculate the pH and ionization of a 1.75 M $NH_3$ solution in 2.50 M $NH_4Cl$?
Ammonia is a weak base and ionises in water: $sf(NH_3+H_2OrightleftharpoonsNH_4^++OH^-)$ For which: $sf(K_b=([NH_4^+][OH^-])/([NH_3])=1.8xx10^(-5)color(white)(x)"mol/l")$ These are equilibrium concentrations. The ICE table in mol/l is: $" "sf(NH_3"...
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