Question

For the following reaction, 22.0 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas . nitrogen monoxide ( g ) + hydrogen ( g ) nitrogen ( g ) + water ( l ) What is the maximum amount of nitrogen gas that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete

What is a limiting reagent?

The reactant that is entirely used up in a reaction is called as limiting reagent.

The reaction:

$2NO(g) +2H_2(g)$ → $N_2 +2H_2O$

Moles of nitrogen monoxide

Molecular weight: $M_(_N_O_)$=30g/mol

$n_(_N_O_) =\frac{mass}{molar \;mass}$

$n_(_N_O_) =\frac{22.0}{30g/mol}$

$n_(_N_O_) = 0.73 mol$

Moles of hydrogen

Molecular weight: $M_(_H_2_)$=30g/mol

$n_(_H_2_) =\frac{mass}{molar \;mass}$

$n_(_H_2_) =\frac{5.80g}{2g/mol}$

$n_(_H_2_) = 2.9 mol$

Hydrogen gas is in excess.

NO is the limiting reagent.

The amount of the excess reagent remains after the reaction is complete.

$n_(_N_2_) =$ (2.9 mol- 0.73 mol NO x $\frac{1 \;mol \;of \;H_2}{2 \;mole \;of \;NO}$) x $\frac{2g \;of \;H_2}{mole \;of \;H_2}$

$n_(_N_2_) =$4.34 g

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