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have you got an equation for the reaction

no

do you know the for mula for ammonia?

is it NH_3 ?

nh3

so far we have
\[N_2+H_2 \longrightarrow NH_3\]
right?

can you balance the equation for number of H and N on each side?

UR EQ IS NT APPEARING

oh,

N_2 + H_2 ⟶ NH_3

so to make the product we need thee time as much H as N

Is this always the case??

ITS AGAIN APPEARING MATHS PROCCESING ERROR

i think ujjwal is right

N2 + 3H2 ---> 2NH3

which is the limiting reagent ?

Obviously, hydrogen is in excess and nitrogen is the limiting reagant

so according to the balanced equation the theoretical yield will be 2 moles of NH_3

yep! 2 mol of NH3 will be yielded!

the molar mass of ammonia is (1x14+3x1) g/mol

can you finish @Ruchi. ?

Where are u stuck @Ruchi.?

Did u understand the part where the moles are calculated?