Chemistry question help

SIMpLiFy SG

In an experiment,excess zinc powder reacts with 50cm cube of 0.1dm cube/mol sulphuric acid in an experiment.The products formed in the reaction are zinc sulphate and hydrogen gas.

(a)Write a balanced equation for the reaction.
(b)Determine the volume of hydrogen gas produced in the reaction.

Thanks.

SBS3625D SG

WTF.

You might as well give me a table, a calculator and your exam paper?

Do it yourself please....there's only so much we can help you.

wishboy SG

a)
Zn (s) + H2SO4 (aq) ----> ZnSO4 (aq) + H2 (g)

b)
No. of mol H2SO4
= 0.05/0.1
= 0.5 mol

Vol. of H2
= 0.5*24
= 12 dm³

i not sure whether my answers correct anot
anyway try to understand how to do
anything dunno ask ur teacher/classmate/anyone who noes how to do

SIMpLiFy SG

ok..sry , i just noticed my teacher havn’t taught this topic yet , nowonder it looks so unfamiliar..well thx anyway =D.

gnab SG

6Zn (s) + 2HSO4 (aq) -> 2Zn3SO4 (s) + H2 ( g)

no of mol of HSO4 used = (50/1000)/.1 = 0.5mol
no of mol of H2 produced = no of mol of HSO4 = 0.5mol

hope my answer is useful..

Mr. Green

wishboy SG

er
how come in ur equation sulphuric acid is HSO4? not H2SO4 meh?

BaoBei92 SG

er
how come in ur equation sulphuric acid is HSO4? not H2SO4 meh?

6Zn (s) + 2HSO4 (aq) -> 2Zn3SO4 (s) + H2 ( g)

he used 2HSO4. probably typo.

Darkness_hacker99 SG

This topic is about Limiting Reactant and Chemical Calculation. Limiting reacting as in 'not in excess'.

First must find out which is the limiting reactant, before you can do the calculation. Since zinc power is in excess, then sulpuhruic will be the limiting reactant.

Next is to find the no. of mol of sulphuric acid. With that, you can then calculate volume of hydrogen gas produced.

Darkness_hacker99 SG

It's H₂SO₄ at O Level.

Darkness_hacker99 SG

Your answer correct.

weewee SG

Why is it *24?

Holy i have forgotten my chemistry.

weewee SG

now i remember the formula 1 mole = 24dm3 but forgot why.

Darkness_hacker99 SG

Molar Gas Volume = 24dm³

It's the amount of volume occupied by 1 mole of gas molecule @ r.t.p

eagle SG

now i remember the formula 1 mole = 24dm3 but forgot why.

0.5 mole of H2SO4 still consists of 0.5 mole of H2.

So volume of H2 produced from 0.5 mol of H2SO4 is still 0.5 mol of H2 (because it is by number of molecules).

Since 1 mole = 24dm3, 0.5 mol = 12 dm3 of H2