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
Answers
Answered by
26
answer : option (4) 200g
explanation : when calcium carbonate is heated , gives carbon dioxide and calcium oxide.
i.e.,
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
Answered by
5
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.
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