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3 years ago

What is the mole ratio of D to A in the generic chemical reaction? 2A+B >C+3D

Chemistry

2 answers:

DerKrebs [107]3 years ago

8 0

Answer: D : A is 3 : 2.

Explanation:

The law of definite proportion is being used here. The law states that a given compound always has the same proportion of its constituent elements by mass. The numbers written in the equation are used to balance it, which is an important part whilst determining the mole ratio

Keeping in mind law:

2a + b => c + 3d

Using the concept of stoichiometry coefficients of d and a

So the ratio of d:a is 3:2

Anit [1.1K]3 years ago

7 0

Answer: The mole ratio of D : A is 3 : 2.

Explanation: Mole ratio for a chemical reaction is the ratio of their respective stoichiometric numbers or the ratio of their moles in a chemical reaction.

For a given chemical reaction:

$2A+B\rightarrow C+3D$

By stoichiometry,

2 moles of A is reacting with 1 mole of B to produce 1 mole of C and 3 moles of D.

So, Mole ratio of D is to A will be 3 : 2

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mario62 [17]

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3 years ago

What is true about a saturated solution

ipn [44]

The rate of crystallizing is equivalent to the rate of dissolving.

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4 years ago

A mixture of NaBrO 3 , NaBrO3, NaHCO 3 , NaHCO3, Na 2 CO 3 , Na2CO3, and NaBr NaBr was heated, producing H 2 O , H2O, CO 2 , CO2

nata0808 [166]

Answer:

Composition of initial mixture is:

9.02g of NaBrO₃

15.84g of Na₂CO₃

17.06g of NaHCO₃

82.58g NaBr

Explanation:

For the reactions:

2NaBrO₃(s) ⟶ 2NaBr(s) + 3O₂(g)

2NaHCO₃(s) ⟶ Na₂O(s) + H₂O(g) + 2CO₂(g)

Na₂CO₃(s) ⟶ Na₂O(s)+CO₂(g)

All H₂O(g) comes from NaHCO₃. Thus, initial moles and mass of NaHCO₃ are:

1.83g H₂O ₓ (1 mol H₂O / 18.02g) ₓ (2 mol NaHCO₃ / 1 mol H₂O) = 0.203moles NaHCO₃ ₓ (84g / 1mol NaHCO₃) =

17.06g of NaHCO₃

CO₂ comes from NaHCO₃ and Na₂CO₃.

15.51g of CO₂ are:

15.51g CO₂ ₓ (1mol / 44.01g) = 0.352moles of CO₂

As 2 moles of NaHCO₃ produce 2 moles of CO₂, moles of CO₂ that comes from NaHCO₃ are 0.203moles NaHCO₃. Moles of CO₂ that comes from Na₂CO₃ are:

0.352mol CO₂ - 0.203mol CO₂ = 0.149mol CO₂

These moles of CO₂ are produced from:

0.149mol CO₂ ₓ (1 mol Na₂CO₃ / 1 mol CO₂) ₓ (106g / 1mol Na₂CO₃) =

15.84g of Na₂CO₃

And all O₂ comes from NaBrO₃. Initial mass of NaBrO₃ is:

2.87g O₂ ₓ (1 mol O₂ / 32g) ₓ (2 mol NaBrO₃ / 3 mol O₂) ₓ (150.9g / 1mol NaBrO₃) =

9.02g of NaBrO₃

If initial mass of the mixture was 124.5g, mass of NaBr was:

124.5g - 9.02g of NaBrO₃ - 15.84g of Na₂CO₃ - 17.06g of NaHCO₃ =

82.58g NaBr

Composition of initial mixture is:

9.02g of NaBrO₃

15.84g of Na₂CO₃

17.06g of NaHCO₃

82.58g NaBr

5 0

4 years ago

A chemist adds 405.0mL of a 0.79M sodium carbonate solution to a reaction flask. Calculate the millimoles of sodium carbonate th

Sophie [7]

Answer:

320 mmol

Explanation:

405.0 mL * 1L/1000 mL = 0.4050 L

0.79 M = 0.79 mol/L

0.79 mol/L * 0.4050 L=0.32 mol

0.32 mol * 10^3 mmol/1 mol = 320 mmol

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4 years ago

Using a chemical equation to find moles of product from moles...

kaheart [24]

Answer: 0.600 moles

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2 NH4ClO4 (s) ----> N2 (g) + Cl2 (g) + 2 O2 (g) + 4 H2O (g)

Since, 2 moles of NH4ClO4 produces 2 moles of O2 gas, this means that 0.6 moles of NH4ClO4 will produce 0.6 moles of O2 gas

7 0

2 years ago