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Paha777 [63]
3 years ago
12

using the equation 2H2+O2-->2H2O,if 192g of water are produced ,how many grams of hydrogen must react with it

Chemistry
1 answer:
Sveta_85 [38]3 years ago
7 0
Based on the given chemical reaction, it can be said that the equation is balanced. To get the amount of Hydrogen needed, back calculation can be done. From 192 g of water, it can be related to its molar mass to convert to the amount of mol. Based on the equation, 1 mol of H2 is equivalent to 2 mol water, by this the amount of H2 needed can be calculated by multiplying to the molar mass (18 g/mol). Below is the calculation:
192 g H2O*(1 mol H2O/18 g H2O)(1 mol H2/2 mol H2O)*(2 g H2/1 mol H2) = 10.67 g H2
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2 years ago
What makes the shadow during a solar eclipse
Damm [24]
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3 years ago
A sample of octane undergoes combustion according to the equation 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH°rxn = -11018 kJ. What mas
Anna71 [15]

Answer:

\large \boxed{\text{528.7 g} }

Explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be 11018 "moles" of "kJ"  

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:                      32.00

              2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ

n/mol:                                                                  7280

1. Moles of O₂

The molar ratio is 25 mol O₂:11 018 kJ

\text{Moles of O}_{2} = \text{7280 kJ} \times \dfrac{\text{25 mol O}_{2}}{\text{11 018 kJ}} = \text{16.52 mol O}_{2}

2. Mass of O₂

\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}

3 0
3 years ago
Help me answer this chemistry question. thank you!
Vladimir79 [104]

Answer:

Chlorine and bromine

Explanation:

I think thats the answer in your question

8 0
3 years ago
Calculate the volume of the acid solution and the volume of the conjugate base solution that would be needed to prepare a buffer
bogdanovich [222]

Answer:

Explanation:

This can be contradictory, depending on whether the 0.1 M

is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

VA− = 9.125 mL

VHA = 15.875 mL

The Henderson-Hasselbalch equation is:

pH = pKa + log [A−][HA]

We have a pH 4.5

solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

[A−][HA]=10

pH − pKa = 104.5 − 4.74 = 0.5754

Now, if the total concentration is

0.10 M , then:

[HA] + [A−] 0.5754

[HA] = 0.10 M

⇒[HA] = 0.10 M 1.0000 +0.5754

= 0.0635 M

−−−−−−−−

⇒[A−] = 0.0365 M

−−−−−−−−

and these concentrations are AFTER mixing. Since the total volume is 50 mL , or 0.050 L, the mols of each component (which are constant!) are:

nA − = 0.0365 molL × 0.050L =

0.001825 mols

−−−−−−−−−−−−

nHA = 0.0635 molL × 0.050L =

0.003175 mols

−−−−−−−−−−−−

So, if both of the starting concentrations were

0.20 M, we can find the volume they each start with:

VA − = 1 L0.20mols

A− × 0.001825mols A− = 0.009125 L = 9.125 mL

−−−−−−−−

VHA = 1 L 0.20 mols HA × 0.003175

mols HA = 0.015875 L = 15.875 mL

−−−−−−−−−

And this should make sense, because the total starting volume is

25.000 mL , the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.

6 0
3 years ago
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