Question #89a6e

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Question #89a6e

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Answer 1 · 2018-01-31T21:07:55

$a . N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ $a. N_2(g)+3H_2(g)->2NH_3(g)$
$b . \approx 577 g N H_{3}$ $b. ~~577gNH_3$
$c . \approx 69.0 % y i e l d$ $c. ~~69.0% yield$

Explanation:

Write and balance the given chemical equation.
$N_{2} + 3 H_{2} . \to 2 N H_{3}$ $color(red)(N_2+3H_2.->2NH_3$
Find the required molar masses of the involved substances.
$N_{2} = 28.0 g/mol$ $N_2=28.0" g/mol"$
$N H_{3} = 17.0 g/mol$ $NH_3=17.0" g/mol"$
Given the masses of both raw material and the production output, theoretical yield of this production run can be computed as follows:
$=475cancel(gN_2)xx(1molN_2)/(28.0cancel(gN_2))$
$=16.96molN_2~~17.0molN_2$

Now, convert mole $O_2$ $(etaO_2)$ to $etaNH_3$ using the molar ratio of these two(2) substances that is obtainable from the balanced equation described in $ul ("step " 1)$ .
$=17.0cancel(molN_2)xx(2molNH_3)/(1cancel(molN_2))$
$=33.93molNH_3~~34.0molNH_3$

Then, find the $"theoretical yield"$ . Knowing the molar mass of $NH_3$ , conversion factor is obtainable and the production output can be computed. Make sure units work out and the desired unit is attained; i.e.,
$color(red)(=34.0cancel((molNH_3)xx(17.0gNH_3)/(1cancel(molNH_3))$
$color(red)(=576.79gNH_3~~577gNH_3)$

Finally, find the percent yield in this production run; i.e.,
$% yield=("actual yield")/("theoretical yield")xx100$
$% yield=(397cancel(gNH_3))/(577cancel(gNH_3))xx100$
$color(red)(% yield=68.8%~~69.0%$

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Question #89a6e

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