Question

A student determines that according to the reaction N2(g) + 3H2(g) --> 2NH3(g), if 34g of nitrogen gas is reacted with excess hydrogen, 41g of ammonia can be produced. When the actual reaction was complete, only 38g of ammonia formed. Determine the percent yield for the reaction.

Answer 1

Answer:

The percent yield of this reaction is 92.7 %

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 34.0 grams

Mass of ammonia (NH3 produced = 41.0 grams

Molar mass of N2 = 28.0 g/mol

Molar mass of NH3 = 17.02 g/mol

Actual yield of ammonia = 38 grams

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles N2 = 34.0 grams / 28.0 g/mol

Moles N2 = 1.214 moles

Step 4: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 1.214 moles N2 we'll have 2* 1.214 = 2.428 moles NH3

Step 5: Calculate mass NH3

Mass NH3 = moles * molar mass

Mass NH3 = 2.428 moles * 17.02 g/mol

Mass NH3 = 41 grams

Step 6: Calculate percent yield for the reaction

Percent yield = (actuald yield / theoretical yield) * 100 %

Percent yield = (38 grams / 41 grams ) * 100 %

Percent yield = 92.7 %

The percent yield of this reaction is 92.7 %

Answer 2

Answer:

$Y=92\%$

Explanation:

Hello,

In this case, by considering the given chemical reaction, with given mass of nitrogen, one could compute the theoretical yield of ammonia as shown below and considering their 1 to 2 molar relationship in the chemical reaction:

$m_{NH_3}^{theoretical}=34gN_2*\frac{1molN_2}{28gN_2}*\frac{2molNH_3}{1molN_2}*\frac{17gNH_3}{1molNH_3} \\m_{NH_3}^{theoretical}=41.3gNH_3$

In such a way, the percent yield is obtained as shown below:

$Y=\frac{m_{NH_3}^{actual}}{m_{NH_3}^{theoretical}} *100\%=\frac{38g}{41.3g} *100\%\\\\Y=92.0\%$

Best regards.