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Over [174]
3 years ago
11

suppose that 0.05 g of NaHCO3 reacts with 0.05 g of HCI and the reaction bubbles due to the carbon dioxide being produced. if th

e substance produced are 0.04 g water and 0.03g of salt how many grams if carbon dioxide gas were produced
Chemistry
1 answer:
VLD [36.1K]3 years ago
6 0

The balanced chemical equation for the given reaction:

NaHCO₃ + HCl --> NaCl + H₂O + CO₂

Given, mass of NaHCO₃ = 0.05 g

Molar mass of NaHCO₃ = 84.007 g/mol

Moles = mass/molar mass

Moles of NaHCO₃ = 0.05 g/84.007 g/mol

Moles of NaHCO₃ = 6.0 x 10⁻⁴

Given, mass of HCl = 0.05 g

Molar mass of HCl = 36.46 g/mol

Moles of HCl = 0.05 g/ 36.46 g/mol

Moles of HCl = 1.4 x 10⁻³

Since NaHCO3 has the least number of moles, it is the limiting reagent.

As per the balanced equation the molar ratio between NaHCO₃ : CO₂ is 1:1

Therefore, moles of CO₂ = 6.0 x 10⁻⁴

Molar mass of CO₂ = 44 g/mol

Mass = moles x molar mass

Mass of CO₂ =  6.0 x 10⁻⁴ mol x 44 g/mol = 0.0264 g

Mass of carbon dioxide gas produced = 0.0264 g

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Two molecules of one reactant combine with 3 molecules of another to produce 5 molecules of a product.
maksim [4K]

Answer:

A = 2A + 3B → 5C

Explanation:

The two molecule of A and three molecules of B will react to form the five molecules of C.

2A + 3B   →   5C

Other options are incorrect because,

B = A₂ + B₃  →   C₅

in this reaction one molecule of A₂ and one molecule of B₃ combine to form one molecule of C₅.

C = 2A + 5B   →  3C

in this reaction two molecules of A and five molecules of B combine to form three molecule of C.

D = A₂ + B₃  →  C₃

in this reaction one molecule of  A₂ and one molecule of B₃ combine to from one molecule of C₃.

3 0
3 years ago
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
Hi can u help me pls? I'm totally stuck . The natural source of acidity in rain water is _____.
kykrilka [37]

Answer-The correct option is option d with says all of the above.

Explanation- All three acids that are given combined together to form acid rain in which nitric and sulphuric acid are stronger acids present while carbonic acid is a weaker one.

The carbon dioxide admitted in air combines with water to form carbonic acid and gives a weak acidic nature to rainwater. Pollution in nature makes sulphur and nitrogen present in air react to form the stronger acids responsible for acid rain.

5 0
3 years ago
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
3 years ago
Why does magma composition change during fractional crystallization?.
katrin [286]
Why does magma composition change during fractional crystallization? Different elements in the magma form crystals at different rates, leaving behind more of the unused elements. ... The crystals are denser than the magma.
6 0
2 years ago
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