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2 years ago

How many milliliters of a 0.8 M solution of citric acid would be needed to react with 15 grams of baking soda? Show your work.

Chemistry

1 answer:

Alinara [238K]2 years ago

5 0

Answer:

75mL

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

C6H8O7 + 3NaHCO3 → Na3C6H5O7 + 3CO2 + 3H2O

Step 2:

Conversion of 15g of baking soda (NaHCO3) to mole.

This is illustrated below:

Mass of NaHCO3 = 15g

Molar Mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 23 + 1 + 12 + 48 = 84g/mol

Number of mole = Mass/Molar Mass

Number of mole of NaHCO3 = 15/84 = 0.179 mole

Step 3:

Determination of the number of mole of citric acid (C6H8O7) produced by the reaction. This is illustrated below:

From the balanced equation above,

1 mole of C6H8O7 reacted with 3 moles of NaHCO3.

Therefore, Xmol of C6H8O7 will react with 0.179 mole of NaHCO3 i.e

Xmol of C6H8O7 = 0.179/3

Xmol of C6H8O7 = 0.06 mole.

Step 4:

Determination of the volume of C6H8O7 needed for the reaction. This is illustrated below:

Molarity of C6H8O7 = 0.8 M

Mole of C6H8O7 = 0.06

Volume =..?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 0.06/0.8

Volume = 0.075L

Converting 0.075L to mL, we have:

1L = 1000mL

Therefore 0.075L = 0.075 x 1000 = 75mL.

Therefore, 75mL of citric acid (C6H8O7) is needed for the reaction.

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Please help me double-check my answer: Calculate the molarity of an aqueous solution that contains 36.5g KMnO4 and has a total v

Helen [10]

Answer:

The answer to your question is Molarity = 0.6158, I got the same answer as you.

Explanation:

Data

Molarity = ?

Mass of KMnO₄ = 36.5 g

Total volume = 375 ml

Process

1.- Calculate the Molar mass of KMnO₄

KMnO₄ = (1 x 39.10) + (54.94 x 1) + (16 x 4)

= 39.10 + 54.94 + 64

= 158.04 g

2.- Calculate the moles of KMnO₄

158.04 g of KMnO₄ ------------------- 1 mol

36.5 g of KMnO₄ --------------------- x

x = (36.5 x 1) / 158.04

x = 0.231 mol

3.- Convert the volume to liters

1000 ml -------------------- 1 L

375 ml --------------------- x

x = (375 x 1)/1000

x = 0.375 L

4.- Calculate the Molarity

Molarity = moles / volume

-Substitution

Molarity = 0.231 moles / 0.375 L

Result

Molarity = 0.6158

6 0

2 years ago

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A sample of 508.4 grams of copper completely reacted with oxygen to form 572.4 grams of a copper oxide product. how many grams o

Svet_ta [14]

According to law of conservation of mass, mass can neither be destroyed nor created in a chemical reaction. Thus, sum of masses of reactants must be equal to sum of masses of products in a reaction.

The chemical reaction is as follows:

$2Cu+O_{2}\rightarrow 2CuO$

Here, sum of masses of Cu and oxygen gas should be equal to CuO formed.

$2m_{Cu}+m_{O_{2}}=2m_{CuO}$

Thus, mass of oxygen will be:

$m_{O_{2}}=2(572.4-508.4)g=128 g$

This can be further proved as follows:

The balanced chemical reaction is as follows:

$2Cu+O_{2}\rightarrow 2CuO$

Here, 2 moles of Cu completely reacts with 1 mole of $O_{2}$ to give 2 moles of $CuO$ .

Thus, 1 mole of Cu reacts with 0.5 moles of $O_{2}$ .

The mass of Cu is 508.4 and molar mass is 63.546 g/mol, number of moles can be calculated as follows:

$n=\frac{m}{M}=\frac{508.4 g}{63.546 g/mol}=8 mol$

Thus, number of moles of $O_{2}$ reacting will be:

$n_{O_{2}}=8\times 0.5 mol=4 mol$

Molar mass of oxygen molecule is 32 g/mol thus, mass can be calculated as follows:

m=n×M=4 mol×32 g/mol=128 g/mol

This satisfies the law of conservation of mass.

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2 years ago

A sample of a compound containing only carbon and oxygen decomposes and produces 24.50g of carbon and 32.59g of oxygen. what is

Fynjy0 [20]

<span>Carbon Monoxide. First, determine the relative number of moles of each element by looking up the atomic weights of carbon and oxygen Atomic weight carbon = 12.0107 Atomic weight oxygen = 15.999 Moles of Carbon = 24.50 g / 12.0107 g/mol = 2.039847802 mol Moles of Oxygen = 32.59 g / 15.999 g/mol = 2.037002313 mol Given that the number of moles of both carbon and oxygen are nearly identical, it wouldn't be unreasonable to think that the empirical formula for the compound is CO which also happens to be the formula for Carbon Monoxide.</span>

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2 years ago

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There are two types of nucleic acids, dna and rna. nearly all organisms use dna, not rna, as the central repository for genetic

Leya [2.2K]

Found the choices. Pls see attachment.

The statements that explains this phenomenon are:
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2 years ago

How many grams of NH3 can be prepared from the synthesis of 77.3 grams of nitrogen and 14.2 grams of hydrogen gas?

lbvjy [14]

Answer:

80.41 g

Explanation:

Data Given:

Mass of Nitrogen (N₂) = 77.3 g

Mass of Hydrogen (H₂) = 14.2 g

many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

N₂ + 3 H₂ ------—> 2 NH₃

1 mol 3 mol

As 1 mole of Nitrogen react with 3 mole of hydrogen

Convert moles to mass

molar mass of N₂ = 2(14) = 28 g/mol

molar mass of H₂ = 2(1) + 2 g/mol

Now

N₂ + 3 H₂ ------—> 2 NH₃

1 mol (28 g/mol) 3 mol(2g/mol)

28 g 6 g

28 grams of N₂ react with 6 g of H₂

if 28 grams of N₂ produces 6 g of H₂ so how many grams of N₂ will react with 14.2 g of H₂.

Apply Unity Formula

28 g of N₂ ≅ 6 g of H₂

X g of N₂ ≅ 14.2 g of H₂

Do cross multiply

X g of N₂ = 28 g x 14.2 g / 6 g

X g of N₂ = 66.3 g

As we have given with 77. 3 g of N₂ but from this calculation we come to know that 66.3 g will react with 14.2 g of hydrogen and the remaining 10 g N₂ will be in excess

So, Hydrogen is limiting reactant in this reaction and the amount of NH₃ depends on the amount of hydrogen.

Now

To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

N₂ + 3 H₂ ------—> 2 NH₃

6 g 2 mol

So, 6 g of hydrogen gives 2 moles of NH₃, then how many moles of NH₃ will be produce by 14.2 g

Apply Unity Formula

6 g of H₂ ≅ 2 mol of NH₃

14.2 g of H₂ ≅ X mol of NH₃

Do cross multiply

X mol of NH₃= 14.2 g x 2 mol / 6 g

X mol of NH₃ = 4.73 mol

So, 14.2 g of hydrogen gives 4.73 moles of NH₃

Now

Convert moles of NH₃ to mass

Formula will be used

mass in grams = no. of moles x molar mass . . . . . . (2)

Molar mass of NH₃

Molar mass of NH₃ = 14 + 3(1)

Molar mass of NH₃ = 14 + 3 = 17 g/mol

Put values in equation 2

mass in grams = 4.73 mole x 17 g/mol

mass in grams = 80.41 g

mass of NH₃= 80.41 g

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2 years ago