Question

I need help with 2nd and 11th. It's not urgent, take your time! (:

Answer 1

☆General things:

▪︎the amount of products formed and the amount of reactants consumed are based upon the multiple of their stiochiometric coefficient.

▪︎Limiting reagent(LR): It is that part of reactant which will be first get consumed. After its consumption the reaction stops.

That means the amount by which the products formed will be depend on the limiting reagent consumption.

》For finding LR = moles/stiochiometric coef.

one which has the lowest of its value will be LR in the rxn.

☆solution:

2) $\ 2Al(OH)_{3} + 3H_{2}SO_{4} \rightarrow{ Al_{2}SO_{4} + 6H_{2}O}$

•lets take the Moles of $\ 2Al(OH)_{3}$ as 'a' and $\ 3H_{2}SO_{4}$ as 'b'.

• after the reation ...these will the moles:

1. $\ 2Al(OH)_{3}$ = a-2x
2. $\ 3H_{2}SO_{4}$ = b-3x
3. $\ Al_{2}SO_{4}$ = x
4. $\ 6H_{2}O$ = 6x

》we can observe here that moles of water formed is 6 times the moles of aluminium sulphate.

》x= 1.6 (given in ques)

》moles of water = 6×1.6 = 9.6 moles.

---------

11) In the reaction, by using LR formula, you'll see that $\ H_{2}C_{2}O_{4}$ will be the limiting reagent in ques.

》so products will be formed acc to it.

》moles of $\ 2MnCl_{2}$?

• let's use unitary method:

• 5 moles of $\ H_{2}C_{2}O_{4}$ ----- 2 moles of $\ 2MnCl_{2}$

• 1 mole of $\ H_{2}C_{2}O_{4}$ ----- 2/5 moles of $\ 2MnCl_{2}$

• 3 moles of $\ H_{2}C_{2}O_{4}$ ----- $\dfrac{2×3}{5}$ moles of $\ 2MnCl_{2}$

》moles of $\bf{ 2MnCl_{2}}$ = 1.2

-----

See the attachment! for any problem in understanding.

hope its correct !

Good luck ♧♧

Answer 2

photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.