Question

Hello brainlians !!
Today's Question:-
$\sf \bf \: How \: much \: hydrogen \: and \: nitrogen \: will \: be \: required \: to \: produce \: 85 \: g \: of \: \:ammonia ? \:$
$\underline {\boxed{ \sf{ \purple{Equation: H_2 + N_2 \ \ \ ^{iron(catalyst)}_ { \leftarrow^\rightarrow} \: NH_ 3}}}}$
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Answer 1

Answer :

• Amount of N₂ present = 70 g
• Amount of H₂ present = 15 g

Solution:

Step 1 : Balance equation

• 3H₂ + N₂ = 2NH₃

Here,

Three moles of hydrogen reacts with one mole of nitrogen to give 2 moles of ammonia.

Step II: Find moles of ammonia

Given data,

• Mass of ammonia= 85 g
• Molar mass of ammonia = N +3 H = 14 + 3 = 17 g /mol

$\to \sf{Moles \:of \:NH_3 =\dfrac {mass \:of \: NH_3 }{Molar \:mass \:of \: NH_3} }$

$\to \sf{Moles \:of \:NH_3 =\dfrac {85}{17} }$

$\to \sf{Moles \:of \:NH_3 =5 }$

Step III:

$\to \sf{\dfrac {Moles\: of\: recatants }{stoichiometry} =\dfrac {Moles\: of \: products }{stoichiometry} }$

$\to \sf{\dfrac {Moles \:of\: H_2 }{3} = \dfrac {Moles \:of\: N_2 }{1} =\dfrac {Moles \:of\: NH_3 }{2} }$

From the above equation,

• Moles of H₂ = 5 x 3 / 2 = 7.5 moles
• Moles of N₂ = 5 / 2 = 2.5 moles

Step IV : Find the mass of N₂ and H₂

$\to \sf{Moles \:of \:H_2 =\dfrac {mass \:of \: H_2}{Molar \:mass \:of \: H_2} }$

$\to \sf{mass \:of \: H_2 = 7.5 x 2 }$

$\boxed{\sf{mass \:of \: H_2 = 15 g }}$

Similarly,

$\to \sf{mass \:of \: N_2 = 2.5 x 28 }$

$\boxed{\sf{mass \:of \: N_2 = 70 g }}$

Answer 2

Amount of N₂ present = 70 g

Amount of H₂ present = 15 g

Solution:

Step 1 : Balance equation

3H₂ + N₂ = 2NH₃

Here,

Three moles of hydrogen reacts with one mole of nitrogen to give 2 moles of ammonia.

Step II: Find moles of ammonia

Given data,

Mass of ammonia= 85 g

Molar mass of ammonia = N +3 H = 14 + 3 = 17 g /mol

Moles of NH₃ = mass of NH₃/molar mass of NH₃

Moles of NH₃ = 85/17 = 5