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KiRa [710]
3 years ago
7

Work the entire problem from the beginning. How many grams of H2 would be formed if 34 grams of carbon reacted with an unlimited

amount of H2O? The reaction is: C + H2O → CO + H2 The atomic mass of C is 12.01 g/mole. The atomic mass of H2 is 2.016 g/mole. grams of hydrogen
Chemistry
2 answers:
Alik [6]3 years ago
3 0
34 grams of carbon is approx. 2.833 moles(because molar mass of carbon is 12 grams). Since there is unlimited water, Carbon is the limiting reagent, so taking into account that the equation is already balanced and the fact that in the balanced equation the mole ratio between carbon and Hydrogen gas is 1 to 1, there are 2.833 moles of hydrogen, multiply that by the molecular mass of the Hydrogen gas you get 5.71 grams of Hydrogen gas.
Liula [17]3 years ago
3 0

Answer: 5.66 g of H_2 would be formed if 34 grams of carbon reacted with an unlimited amount of H_2O

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

\text{Number of moles of carbon}=\frac{34g}{12g/mol}=2.83moles

C+H_2O\rightarrow CO+H_2

Carbon is the limiting reagent as it limits the formation of product and hydrogen is the excess reagent.

According to stoichiometry :

1 mole of carbon gives = 1 mole of H_2

Thus 2.83 moles of carbon gives =\frac{1}{1}\times 2.83=2.83moles of H_2

Mass of H_2=moles\times {\text {Molar mass}}=2.83moles\times 2g/mol=5.66g

Thus 5.66 g of H_2 would be formed if 34 grams of carbon reacted with an unlimited amount of H_2O

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3 0
2 years ago
What is the molar solubility of marble (i.e., [ca2 ] in a saturated solution in normal rainwater, for which ph=5.60? express you
Brut [27]
Missing in your question :

Ksp of(CaCO3)= 4.5 x 10 -9

Ka1 for (H2CO3) =  4.7 x 10^-7

Ka2 for (H2CO3) = 5.6 x 10 ^-11

1) equation 1 for Ksp = 4.5 x 10^-9 

CaCO3(s)→ Ca +2(aq)    +  CO3-2(aq)  

2) equation 2 for Ka1 = 4.7 x 10^-7

 H2CO3 + H2O → HCO3- + H3O+

3) equation 3 for Ka2 = 5.6 x 10^-11

 HCO3-(aq) + H2O(l) → CO3-2 (aq)  + H3O+(aq)

so, form equation 1& 2&3 we can get the overall equation:
CaCO3(s)  +  H+(aq)  → Ca2+(aq)   + HCO3-(aq)

note: you could get the overall equation by adding equation 1 to the inverse of equation 3 as the following:
when the inverse of equation 3 is :

CO3-2 (aq) + H3O+ (aq) ↔ HCO3- (aq) + H2O(l)  Ka2^-1 = 1.79 x 10^10
when we add it to equation 1
CaCO3(s) ↔ Ca2+(aq)  +  CO3-2(aq)   Ksp = 4.5 x 10^-9

∴ the overall equation will be as we have mentioned before:
when H3O+ = H+

CaCO3(s) + H+(aq)  ↔ Ca2+ (aq) + HCO3-(aq)   K= 80.55

from the overall equation:

∴K = [Ca2+][HCO3-] / [H+]

when we have [Ca2+] = [HCO3-] so we can assume both = X

∴K = X^2 / [H+]

when we have the PH = 5.6 so we can get [H+]

PH = - ㏒[H+]
5.6 = -㏒[H]
∴[H] = 2.5 x 10^-6

so, by substitution on K expression:

∴ 80.55 = X^2 / (2.5 x10^-6)

∴X = 0.0142

∴[Ca2+] = X = 0.0142 
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