Zn(s) + 2 HCl(aq) —> ZnCl2(aq) + H2(g)
If a scientist wanted to create 10L of hydrogen gas at RTP, what mass of HCl would they need to start with?

First write the balanced equation for the reaction of HCl with Zn (NOTE: the previous answer did NOT do this and thus obtained an incorrect answer)

Zn + 2HCl ==> ZnCl2 + H2(g)

moles of H2 desired:  PV = nRT

n = PV/RT = (1 atm)(10L)/(0.0821 Latm/Kmol)(298K)

n = 0.409 mole H2 desired

moles Zn needed:

0.409 moles H2 x 1 mole Zn/2 moles HCl = 0.204 moles Zn needed

mass Zn needed = 0.204 moles Zn x 65.4 g/mol = 13.4 g Zn needed (to 3 sig. figs.)

Explanation:people always get this wrong but i got you

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.