Question

Please help with stoichiometric mass-to-mass conversions. Question attached

Answer 1

do you know how to do this problem?

Answer 2

\[(3.05 \times 10^3 kg Ca(HCO_3)) \times \frac{ 1000g (Ca(HCO_3)) }{ 1kg (Ca(HCO_3)) } \times \frac{ mole (Ca(HCO_3)) }{ grams (Ca(HCO_3)) } \times \frac{ mole (CaCO_3)}{ mole (Ca(HCO_3)) }\times \frac{ grams (CaCO_3) }{ moles (CaCO_3) }\]

Answer 3

wow thanks so much! just some of the problem got cut off and i cant see some of it.

Answer 4

Sorry, give me a moment and I can retype it. You will just have to fill in the numbers in the appropriote places.

Answer 5

oky thanks soo much!!

Answer 6

From where it cut off:
\[\frac{ mole (Ca(HCO_3)) }{ grams (Ca(HCO_3) } \times \frac{ mole (CaCO_3) }{ grams (CaCO_3) } \times \frac{ grams (CaCO_3) }{ moles (CaCO_3) }\]

Answer 7

If the question wants the answer in kilograms you will have to provide one more step where you convert CaCO_3 from grams to kilograms.

Answer 8

okay this REALLY helps!! thnk you!!

Answer 9

No problem :)

Answer 10

with thegrams of Ca(HCO)3 what would they be?
do you have you add hydrogen carbon and oxygen and then multiply it by 3?

Answer 11

oh wait nevermind. that question made no sense. lol

Answer 12

My final answer is 1.58 times 10^10

Answer 13

the answer in the book says 1.88 times 10^10 kg CaCO3

Answer 14

I got it now as well; I calculated the mass for my CaCO_3 wrong the first time.

Answer 15

oh okay. i am still doing the calculations. lol. these problems are so hard for me!

Answer 16

When I first started doing stoichiometry it was very confusing; \[ \frac{ 1000 grams }{ 1 kilogram }\]
One thing I learned is to always put larger numbers with the single unit (grams) and the one with the double unit (kilograms). I hope you understand what I mean :)