An excess of hydrogen reacts with 14.0 g of $N_{2}$ $N_2$ . What volume of ammonia will be produced at STP?

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Answer 1 · 2016-05-03T13:19:44

$V = 33.6 L$ $V=33.6L$

The reaction between hydrogen and nitrogen is the following:

$3 H_{2} (g) + N_{2} (g) \to 2 N H_{3} (g)$ $3H_2(g)+N_2(g)->2NH_3(g)$

Using the mass of nitrogen $m = 14.0 g$ $m=14.0g$ , we can first find the number of mole of ammonia that will be produced:

$?molNH_3=14.0cancel(gN_2)xx(1cancel(molN_2))/(28.0cancel(gN_2))xx(3molNH_3)/(1cancel(molN_2))=1.50molNH_3$

It is known that at STP, one mole of gas occupies a volume of $22.4L$ . This is called the molar volume.

Thus, the ammonia produced will occupy the following volume:

$?L=1.50cancel(molNH_3)xx(22.4L)/(1cancel(molNH_3))=33.6 L$

An excess of hydrogen reacts with 14.0 g of N_2N2N_2. What volume of ammonia will be produced at STP?