Question

How much calcium carbonate do I need to produce 50mL carbon dioxide? 2HCl + CaCO3 -> CaCl2 + H2O + CO2

*I'm pretty sure I need 90.91mL of HCl... but I could be very wrong.

please help.

Answer 1

we need `0.2234`grams of `CaCO_3`to produce `50`ml of `CO_2`

Explanation:

Given:
`2HCl + CaCO_3 -> CaCl_2 + H_2O + CO_2`
Atomic weight of Hydrogen is 1.0
Atomic weight of Chlorine is 35.5
Atomic weight of Calcium is 40.1
Atomic weight of Carbon is 12.0
Atomic weight of Oxygen is 16.0

Molecular weight of `HCl` is `1(1.0)+1(35.5)=36.5`
Molecular weight of `2HCl` is `2(36.5)=73.0`
Molecular weight of `CaCO_3` is `1(40.1)+1(12)+3(12.0)=100.1`
Molecular weight of `CaCl_2` is `1.0(40.1)+2(35.5)=111.1`
Molecular weight of `H_2O` is `1.0(1.0)+2(16.0)=18.0`
Molecular weight of `CO_2` is `1.0(12.0)+2(16.0)=44.0`

Check
lhs`=73.0+100.1=173.1`
rhs`=111.1+18.0+44.0=173.1`
The equation is balanced
Now,
`44`gms of `CO_2` occupies a volume of `22400`ml at STP

Hence,
`100.1`gms of `CaCO_3`is essential to produce `44.0`gms of `CO_2`
`22400`ml space contains `44`gms of `CO_2`
`100.1`gms of `CaCO_3`is essential to produce `22400`ml of `CO_2`
`x`gms of `CaCO_3`is essential to produce `50`ml of `CO_2`

where `x` indicates amount of calcium carbonate needed to produce `50`ml of `CO_2`
`x:100.1::50:22400`
Product of extremes = Product of means
`x(22400)=100.1(50)`
Solving for `x`
`x=100.1(50/22400)=0.2234`
Thus, we need `0.2234`grams of `CaCO_3`to produce `50`ml of `CO_2`