Question

pls help me


How many liters of NH3 are needed to react completely with 30.0 L of NO (at STP)?4NH3 (g)+ 6NO(g) 5N2(g)+ 6H2o(g)


a.
5.0 L
c.
7.5 L
b.
20.0 L
d.
120.0 L

Answer 1

Answer is: b.20.0 L liters of ammonia are needed to react completely.
V(NO) = 30 L.
n(NO) = V(NO) ÷ Vm.
n(NO) = 30 L ÷ 22,4 L/mol.
n(NO) = 1,34 mol.
From chemical reaction: n(NO) : n(NH₃) = 6 : 4 (3 : 2).
n(NH₃) = 4 · 1,34 mol ÷ 6.
n(NH₃) = 0,892 mol.
V(NH₃) = 0,892 mol · 22,4 L/mol.
V(NH₃) = 20 L.

Answer 2

Answer:

The answer is B. 20 L

Explanation:

Standard temperature and pressure (STP) conditions are conditions used as a reference point for all gases and always the same. Pressure values at 1 atmosphere and temperature at 0 ° C are usually used. The volume of a constant number of moles of gas depends on the temperature and pressure measurements. Therefore, under STP conditions, 1 mole of any gas occupies an approximate volume of 22.4 liters.

With this in mind, you can calculate the amount of moles in 30 L of NO by a simple rule of three: If there are 22.4 L in 1 mole, how many moles are in 30 L?

$\frac{30 L*1 mol}{22.5 L} =1.34 moles$

You know the reaction:  4NH3 (g)+ 6NO(g) ⇒5N2(g)+ 6H2o(g)

By stoichiometry, that is, the amount of reagent so that a certain amount of product is produced, you know that in order to produce 6 moles of NO you need 4 moles of NH3. You now make a simple rule of three: if to produce 6 moles of NO you need 4 moles of NH3, how many moles do you need from the reagent to produce 1.34 moles of NO?

$\frac{1.34moles*4 moles}{6 moles} = 0.89 moles$

Knowing that 1 mole of any gas occupies an approximate volume of 22.4 liters (STP), you calculate what volume 0.89 moles of NH3 occupy: if 1 mole occupies 22.4L, how much volume does 0.89 moles occupy?

$\frac{0.89moles*22.4 L}{1 mol} = 19.94 L$

This is approximately 20 L