Question

Hey, anyone able to help with conversions?

Answer 1

I already know the top answer is wrong :>

Answer 2

100 gm CaCO3 requires 73 gm HCl.

1 gm CaCO3 requires 0.73 gm HCl.

27 gm CaCO3 require 0.73*27 gm HCl I.e; 19.71 gm HCl

Answer 3

But we r given only 10 gm HCl to react with 27 gm CaCO3. So HCl is limiting reagent as it is not present in excess

Answer 4

Now we will calculate the amount of CaCl2 produced with respect to limiting reagent HCl

Answer 5

73 gm HCl produces 111 gm CaCl2.

1 gm HCl produces 111/73 gm CaCl2.

10 gm HCl produces 1110/73 gm CaCl2.

Answer 6

@jiteshmeghwal9 could you help with the second part at all? super stuck still lol

Answer 7

@ganeshie8 @pooja195 either of yall good at chem?:)

Answer 8

111 gm CaCl2 is produced by 100 gm CaCO3.
1 gm CaCl2 is produced by 100/111 gm CaCO3.
1110/73 gm CaCl2 is produced by \(\frac{100*1110}{111*73}\) gm.

Answer 9

The amount of CaCO3 remained after reaction is 27-(amount reacted).

Answer 10

The 27.0 g Calcium Carbonate produces 29.9 g Calcium Chloride.
The 10.0 g HCL produces 15.6 g Calcium Chloride.

So the HCL limits the reaction and produces 15.6 g CaCO3, while the Calcium Carbonate is in excess

Answer 11

The second part you can figure how much CaCO3 will react with 10.0 grams HCl, the difference from the given 27.0 g and that is the excess amount.

Answer 12

14.1 grams CaCO3 is used to react with the 10.0 grams HCl. The remaining amount is
(27.0 - 14.1) grams CaCO3 = 12.9 grams CaCO3