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

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

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

6.67 L

Explanation:

The correct equation is:

 ⇒  

4 moles of reacts completely with 6 moles of NO to give 5 moles of .

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

Hence we can say that :

4L of reacts with 6L of NO.

Then to get 10L of NO L of should react.

=L

the answer is definitely c.

50.23 i believe is correct but don't quote me on that.