You carefully weigh out 14.00 g of CaCO3 powder and add it to 56.70 g of HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 64.96 g . The relevant equation is
CaCO3(s)+2HCl(aq)→H2O(l)+CO2(g)+CaCl2(aq)
Assuming no other reactions take place, what mass of CO2 was produced in this reaction?

Question

You carefully weigh out 14.00 g of CaCO3 powder and add it to 56.70 g of HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 64.96 g . The relevant equation is
CaCO3(s)+2HCl(aq)→H2O(l)+CO2(g)+CaCl2(aq)
Assuming no other reactions take place, what mass of CO2 was produced in this reaction?

raffle_snaffle · Answer

@Photon336

photon336 · Answer

$$CaCO_{3} = 100.1 gramg/mol$$

$$HCl molar mass = 36.46 g/mol $$

$$14.0 g CaCO_{3}*\frac{ mol }{ 100.1grams } = 0.14 moles CaCO_{3}$$

$$56.70g HCl \frac{ mol }{ 36.46 } = 1.6 mol HCL$$

photon336 · Answer

CaCO3 is the limiting reagent so we use that

$$0.14 mol CaCO_{3} *((CO_{2})/(CaCO_{3}) = 0.14 mol CO_{3} * (44g/mol) = 6.2grams$$

raffle_snaffle · Answer

that is wrong. I solved it already. its 5.74g

raffle_snaffle · Answer

suppose to use the law of conservation

raffle_snaffle · Answer

14.00g + 56.70g = 70.70g total

raffle_snaffle · Answer

70.70g - 64.96g = 5.74g lost of CO2

photon336 · Answer

ah .. wait YES. i see.

photon336 · Answer

CO2 bubbled out of your reaction so we did not need to even consider doing stioch to begin with. because the question states that there aren't any side reactions going on.

photon336 · Answer

@raffle_snaffle I see. I had to look at this a second time. In our reaction, CO2 was being released so, that's where the loss of mass was from. technically if we had 70 grams starting we should have the same amount in our products.