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Juliette [100K]

3 years ago

14

How many moles of CO are produced when 35.0 mol C2H2 react completely

1 answer:

zalisa [80]3 years ago

5 0

The answer is 70.0 mol CO2

Mark me the brainliest.

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An industrial synthesis of urea obtains 87.5 kg of urea upon reaction of 68.2 kg of ammonia with excess carbon dioxide. Determin

dolphi86 [110]

Answer:

The theoretical yield of urea = <u>120.35kg</u>

The percent yield for the reaction = <u>72.70%</u>

Explanation:

Lets calculate -

The given reaction is -

$2NH_3(aq)+CO_2$ → $CH_4N_2O(aq)+H_2O (l)$

Molar mass of urea $CH_4N_2O$ = 60g/mole

Moles of $NH_3$ = $\frac{62.8kg/mole}{17g/mole}$ (since $Moles=\frac{mass of substance}{mass of one mole}$ )

= 4011.76 moles

Moles of $CO_2$ = $\frac{105kg}{44g/mole}$

= $\frac{105000g}{44g/mole}$

= 2386.36 moles

Theoritically , moles of $NH_3$ required = double the moles of $CO_2$

but , $4011.76$ , the limiting reagent is $NH_3$

Theoritical moles of urea obtained = $\frac{1 mole CH_4N_2O}{2mole NH_3}\times4011.76 mole NH_3$

= $2005.88mole CH_4N_2O$

Mass of 2005.88 mole of $CH_4N_2O$ = $2005.88 mole \times\frac{60g CH_4N_2O}{1mole CH_4N_2O}$

= 120352.8g

$120352.8g\times \frac{1kg}{1000g}$

= 120.35kg

Therefore , theroritical yeild of urea = 120.35kg

Now , Percent yeild = $\frac{87.5kg}{120.35kg}\times100$

72.70%

Thus , the percent yeild for the reaction is 72.70%

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3 years ago