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2 years ago

How many moles of oxygen are required to make 4 moles of water?

Chemistry

2 answers:

kakasveta [241]2 years ago

8 0

Answer:

2 moles of oxygen

Explanation:

Basic Equation (given) :

2H₂ (g) + O₂ ⇒ 2H₂O (g)

From this, we understand for every 2 moles of water, 1 mole of oxygen is needed

Taking the number of moles of oxygen and water in proportion :

1 : 2
1 × 2 : 2 × 2
2 : 4

Therefore, 2 moles of oxygen are required to make 4 moles of water.

frozen [14]2 years ago

6 0

$\textsf{\qquad\qquad\huge\underline{{\sf Answer}}}$

As per the given equation, we can infer that :

For getting 2 moles mole of water, 1 mole of oxygen gas was required. so to get 4 moles of water we need :

We just have to multiply the whole reaction by 2 on both sides, in order to get 4 moles of H2O on product side.

$\qquad \qquad \rm4 \: H_2 + 2 \: O_2 \rightarrow4 \: H_2O$

therefore, we need

$\qquad \sf \dashrightarrow \: 2 \times 1 = 2 \: moles \: \: oxygen \: gas$

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What is the majority of a solution that contains 0.55 mole of HCl in water to make 1.50 L HCl solution?

PSYCHO15rus [73]

Answer:

0.367M

Explanation:

Molarity refers to the molar concentration of a solution. It can be calculated using the formula below:

Molarity = n/V

Where;

n = number of moles (mol)

V = volume (L)

According to the given information in this question;

n = 0.55 mole

V = 1.50 L

Molarity = 0.55/1.50

Molarity = 0.367M

4 0

3 years ago

Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl

anzhelika [568]

Answer: ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

$CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]$

$\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]$

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Therefore, the enthalpy change for this reaction is, -277.4 kJ

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3 years ago

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3 years ago

The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?

juin [17]

The formula for mole fraction is:

$mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}$ -(1)

The solubility of oxygen gas = 1.0 mmol/L (given)

1.0 mmol/L means 1.0 mmol are present in 1 L.

Converting mmol to mol:

$1.00 mmol\times \frac{1 mol}{1000 mmol} = 0.001 mol$

So, moles of oxygen = 0.001 mol

For moles of water:

1 L of water = 1000 mL of water

Since, the density of water is 1.0 g/mL.

$Density = \frac{mass}{volume}$

$Mass = 1.0 g/ml\times 1000 mL = 1000 g$

So, the mass of water is 1000 g.

Molar mass of water = 18 g/mol.

Number of moles of water = $\frac{1000 g}{18 g/mol} = 55.55 mol$

Substituting the values in formula (1):

$mole fraction = \frac{0.001}{55.55+0.001}$

$mole fraction = 1.8\times 10^{-5}$

Hence, the mole fraction is $1.8\times 10^{-5}$ .

7 0

3 years ago

How many moles of oxygen are required to make 4 moles of water?​

How many moles of oxygen are required to make 4 moles of water?