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Ket [755]
3 years ago
7

Consider a reaction in which hydrogen (H2) and oxygen (O2) react to form water (H2O). By mass, the mixture of reactants is 1 par

t hydrogen and 6 parts oxygen. Write the chemical reaction showing reactant and products including any leftover oxygen or hydrogen. Calculate the equivalence ratio for this reaction.
Chemistry
1 answer:
Murrr4er [49]3 years ago
6 0

Answer:

equivalence ratio= 0.08333

Explanation:

The balanced stoichoimetric equation tells us the amount of hydrogen in moles required to react with a specific amount of oxygen in moles to produce water. The Balanced equation of the reaction is shown

2H_{2}+O_{2}  → 2 H_{2}O

this means 2 parts of Hydrogen will react with one part of oxgen to produce 2parts of water as shown above

in the question, 1 part of hydrogen was made to react with 6parts of oxygen. which means comparing with the balanced equation, oxygen is in excess as only \frac{1}{2} part of oxygen will react with 1part of hydrogen. the equation is shown below

1H_{2}+\frac{1}{2} O_{2}  → 1 H_{2}O +\frac{11}{2}O_{2}  

The equivalence ratio is the ratio of actual fuel to air ratio to stoichiometric fuel to air ratio and is calculated as follows:

actual fuel/air ratio = parts of fuel/parts of air

                               =\frac{1}{6}

stoichiometric fuel/air ratio = \frac{2}{1}

equivalence ratio = \frac{actual fuel/air ratio}{stoichiometric fuel/air ratio }

                             =\frac{\frac{1}{6} }{\frac{2}{1} }

                             = 0.08333

Have a great day Jershaun9358

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What is the mass of 8.12 × 10^23 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)
amid [387]

Answer:

m=59.3gCO_2

Explanation:

Hello,

In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:

M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol

It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:

mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol

Finally, the mass by using the molar mass:

m=1.35mol*\frac{44.009g}{1mol} \\\\m=59.3gCO_2

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8 0
2 years ago
Solution X is a strong base with a pH of 12. Solution X is mixed with solution Y,and the pH of the resulting mixture is 8.Based
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What is the ph of a solution of 1.699 l of 1.25 m hcn, ka = 6.2 x 10-10, and 1.37 moles of nacn?
BlackZzzverrR [31]

The pH of a solution is 9.02.

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8 0
1 year ago
For the following reaction, 5.04 grams of nitrogen gas are allowed to react with 8.98 grams of oxygen gas: nitrogen(g) + oxygen(
OLga [1]

Answer:

1. 10.8 g of NO

2. N₂ is the limting reagent

3. 3.2 g of O₂ does not react

Explanation:

We determine the reaction: N₂(g) + O₂(g) →  2NO(g)

We need to determine the limiting reactant, but first we need the moles of each:

5.04 g / 29 g/mol = 0.180 moles N₂

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Ratio is 1:1, so the limiting reactant is the N₂. For 0.280 moles of O₂ I need the same amount, but I only have 0.180 moles of N₂

Ratio is 1:2. 1 mol of N₂ can produce 2 moles of NO

Then, 0.180 moles of N₂ may produce (0.180 .2) / 1 =  0.360 moles NO

If we convert them to mass → 0.360 mol . 30 g/1 mol = 10.8 g

As ratio is 1:1, for 0.180 moles of N₂, I need 0.180 moles of O₂.

As I have 0.280 moles of O₂, (0.280 - 0.180 ) = 0.100 moles does not react.

0.1 moles . 32 g/mol = 3.2 g of O₂ remains after the reaction is complete.

8 0
3 years ago
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