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Svetach [21]
3 years ago
8

Use the problem below to answer the question: How many grams of H2 would be formed if 34 grams of carbon reacted with an unlimit

ed amount of H2O? The reaction is C + H2O → CO + H2 What substance are we expected to solve for?
Chemistry
2 answers:
laila [671]3 years ago
7 0

Answer : The mass of hydrogen gas H_2 will be, 5.6 grams.

Explanation : Given,

Mass of C = 34 g

Molar mass of C = 12 g/mole

Molar mass of H_2 = 2 g/mole

First we have to calculate the moles of carbon (C).

\text{Moles of }C=\frac{\text{Mass of }C}{\text{Molar mass of }C}=\frac{34g}{12g/mole}=2.8moles

Now we have to calculate the moles of H_2.

The balanced chemical reaction is,

C+H_2O\rightarrow CO+H_2

From the balanced reaction we conclude that

As, 1 mole of C react to give 1 mole of H_2

So, 2.8 moles of C react to give 2.8 moles of H_2

Now we have to calculate the mass of H_2.

\text{Mass of }H_2=\text{Moles of }H_2\times \text{Molar mass of }H_2

\text{Mass of }H_2=(2.8mole)\times (2g/mole)=5.6g

Therefore, the mass of hydrogen gas H_2 will be, 5.6 grams.

stellarik [79]3 years ago
4 0
You have to find how many moles of H2 has formed.

C+ H2O -> CO + H2

mole of carbon reacted = mass / molar mass = 34g/ 12g per mol = 2.83 mol

for every 1 mole of carbon will form 1 mole of H2

mole of hydrogen produced = 1* 2.83 = 2.83 mol

mass of hydrogen = mole * molar mass = 2.83 mol * (1*2) = 5.66g
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Explanation:

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2 years ago
(a) a 0.2 m potassium hydroxide solution is titrated with a 0.1 m nitric acid solution. (i) balanced equation: (ii)what would be
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The question states that the solution is "titrated with a ... nitric acid solution" indicating that \text{HNO}_3 is added to the initially-basic solution. PH value of the solution would keep decreasing as the volume of the acid added increases. The final solution would be acidic as it contains not only water and \text{KNO}_3 (aq), but some \text{HNO}_3 as well. Bromothymol blue would therefore demonstrates a yellow color, the color it present in an acidic solution, at the end of the titration.

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4 0
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