Calculate the volume of h2(g) at 273 k and 2.00 atm that will be formed when 275 ml of 0.725 m hcl solution reacts with 50.0 g zn(s) to give hydrogen gas and aqueous zinc chloride. (r = 0.08206 l atm/k mol)

The volume of  H₂ (g) obtained is 22.4L

Explanation:

First of all, think the reaction:

2HCl (aq) + Zn (s) → ZnCl₂ (aq)  + H₂ (g)

You have to add a 2, in the HCl to get ballanced.

Now we should know how many moles of each reactant, do we have.

Volume . Molarity = moles

Notice that volume is in mL, so I must convert to L.

275 mL = 0.275L

0.275L . 0.725mol/L = 0.2 moles of HCl

Molar mass of Zn: 65.41 g/m

50 g / 65.41 g/m = 0.764 moles

Ratio between reactants is 2:1, so I need the double of moles of HCl to react, and a half moles of Zn to react.

My limiting reactant is the HCl, for 0.764 moles of Zinc, I need 1.528 (0.764 .2) of HCl, and I only have 0.2 moles.

Ratio between HCl and H₂ is 1:1, so 0.2 HCl make 0.2 moles of gas

Now apply the Ideal Gas Law, to find out the volume

P. V = n . R . T

2 atm . V = 0.2 mol . 0.08206L atm/K mol . 273K

V =  (0.2 mol . 0.08206L atm/K mol . 273K ) / 2 atm

V = 2.24 L

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.