Question

question 1 to3, please

Answer 1

Answer:

as per conversion of units,

$1 dm^{3}$ = 1 litre

1) Answer - 0.9 g H2

According to reaction,

         2 mol Al gives 3mol H2

mass of 2 mol Al = 2 x 27 = 54 g

mass of 3 mol H2 = 3 x 2 = 6 g

Now, 54 g Al ------> 6g H2

   So, 8.1g Al -------> $\frac{6 * 8.1}{54}$ = 0.9 g

2) Answer - 1.5 ml

 Same steps as before.

  you will get,

     4mol KO2 ----------> 3mol O2

     284g KO2 ----------->  96g O2

    0.142 g KO2 ----------> $\frac{96 * 0.142}{284}$ = 0.048g O2

Now converting it to vol.

Moles = 0.048 / 32 = 1.5 x $10^{-3}$ moles of O2

1mol = 1L = 1dm^3

So,

1.5 x $10^{-3}$ moles of O2 =  1.5 x $10^{-3}$ L of O2 = 1.5 ml O2

3) Answer - 0.32g CaC2

Same steps but will reverse it.

1 cm^3 = 1ml , so vol. of ethyne gas = 120ml

Moles = $\frac{120ml}{24litre}$ = $\frac{120}{24 000}$ = $\frac{1}{200}$ = 5 x $10^{-3}$ moles of ethyne gas

As per reaction,

   1mol CaC2 ----------> 1mol C2H2 (Ethyne gas)

   So to produce 5 x $10^{-3}$  moles of c2h2 we will need same amount of cac2 = 5 x $10^{-3}$ moles of CaC2

Now, for mass

1mol of CaC2 = 64g (40 + 12+12)

So 5 x $10^{-3}$ mol = 5 x $10^{-3}$ x 64 =   320 x 10^-3 = 0.32 g