Chemistry, 21.12.2019 00:31 mgrvashvi6698

The following balanced equation shows the formation of water.
2h2 + o2 → 2h2o
how many moles of oxygen (o2) are required to react completely with 1.67 mol h2?
a) 0.835 mol o2

b) 1.67 mol o2

c) 3.34 mol o2

Answers: 1

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Answer from: mem8163

Mole ratio of H₂ : O₂ is 2 : 1

∴ if mole of H₂ = 1.67 mol

then mole of O₂ = $\frac{1.67 mol}{2}$

⇒ that mole of oxygen needed to react with H₂ = 0.835 mol

The answer is A

Answer from: dondre54

A) 0.835 mol of O_2

Explanation: According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

As can be seen from the given balanced equation:

$2H_2+O_2\rightarrow 2H_2O$

2 moles of hydrogen are required to react completely with 1 mole of oxygen.

Thus 1.67 moles of hydrogen will react with= $\frac{1}{2}\times 1.67=0.835$ moles of oxygen.

Answer from: Quest

Answerstrong acids and bases completely dissociate in water.this is how strong acids and bases are defined. they will ionize 100% in solution, making a non-reversible reaction

Answer from: Quest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

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The following balanced equation shows the formation of water.
2h2 + o2 → 2h2o
how many...

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