Question

Use the equation below to determine how many liters of NH3 are needed to react completely with 10.0 L of NO (at STP)? Your answer should be in the following format: 1.22 L

4NH3+6NO=5NO+6H2O

Answer 1

6.67 L

Step-by-step explanation:

The correct equation is:

`4NH_(3)+6NO` ⇒
`5N_(2)+6H_(2)O`

4 moles of
`NH_(3)` reacts completely with 6 moles of NO to give 5 moles of
`N_(2)`.

At standard temperature and pressure volume is directly proportional to number of moles of gas.

Hence we can say that :

4L of
`NH_(3)` reacts with 6L of NO.

Then to get 10L of NO
`(4)/(6)*10` L of
`NH_(3)` should react.

=
`(20)/(3)=6.67`L

Answer 2

We need 6.65 L of NH3 to react with 10.0 L of NO

Step-by-step explanation:

Step 1: Data given

volume of NO = 10.0 L

at STP 1 mol has a volume of 22.4 L

Step 2: The balanced equation

4NH3+6NO → 5N2+6H2O

Step 3: Calculate moles of NO

Since 22.4 L = 1 mol

10.0 L = 0.446 mol

Step 4: Calculate moles of NH3

For 6 moles NO, we need 4 moles of NH3 to produce 5 moles of N2 and 6 moles of H2O

for 0.446 moles of NO, we have 4/6 * 0.446 = 0.297 moles of NH3

Step 5: Calculate volume of NH3

1 mol = 22.4 L

0.297 mol = 22.4 * 0.297 = 6.65 L

We need 6.65 L of NH3 to react with 10.0 L of NO