Question

The amount of 25% pure CaCO3 required to
obtain 11.2 litre CO2 at STP when reacts with
H2SO4 is
(1) 100 g
(2) 150 g
(3) 180 g
(4) 200 g​

Answer 1

answer : option (4) 200g

explanation : when calcium carbonate is heated , gives carbon dioxide and calcium oxide.

i.e., $CaCO_3\rightarrow CaO+CO_2$

here it is clear that one mole of calcium carbonate produced one mole of carbon dioxide.

so, amount of calcium carbonate required to produce 11.2litre CO2 at STP= mole of calcium carbonate in 11.2 litre × molar mass of calcium carbonate

[ you should know, volume of 1 mole of gas at STP/ NTP = 22.4L, molar mass of CaCO3 = 100g/mol ]

= 11.2/22.4 × 100

= 50g

but question said “ find 25% pure CaCO3”

so, 25 % of x = 50g

or, 25 × x/100 = 50g

or, x = 200g

hence, amount of 25% pure CaCO3 is 200g

Answer 2

Answer:

GIVEN:

1mole = 22.4 litre

0.5mole=11.2 litre

Explanation:

1mole of Caco3=100gram

0.5 mole=50gram

It is the simplest step ., now

25°/° of pure CaCo3 =50gram

25°/° × X =50gram

25×X/100 = 50

solving, we get,

X= 200.

HOPE IT HELPS YOU.

(^_-)ʕ•ٹ•ʔ(^_^):-)

MARK IT AS A BRAINLIEST ANSWER

THANK YOU

(^o^)丿(^o^)丿(^o^)丿(^o^)丿............