The reaction of hydrochloric acid (HCl) with ammonia (NH3) is described by the equation:

HCl + NH3 → NH4Cl

A student is titrating 50 mL of 0.32 M NH3 with 0.5 M HCl. How much hydrochloric acid must be added to react completely with the ammonia?

A. 6.4 mL

B. 16.0 mL

C. 32.0 mL

D. 50.0 mL

Answer : The correct option is, (C) 32.0 mL

Explanation :

Formula used :

where,

are the initial molarity and volume of NH₃.

are the final molarity and volume of HCl.

We are given:

Now put all the given values in above equation, we get:

Hence, the volume of hydrochloric acid added to react completely with the ammonia must be, 32.0 mL

it is a solid

explanation:

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2