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3 years ago

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

Chemistry

1 answer:

SSSSS [86.1K]3 years ago

6 0

Answer:

There will remain 11.47 grams of Al(OH)3

Explanation:

Step 1: Data given

Mass of sulfuric acid = 35.0 grams

Molar mass sulfuric acid = 98.08 g/mol

Mass of aluminium hydroxide = 30.0 grams

Molar mass of aluminium hydroxide = 78.0 g/mol

Step 2: The balanced equation

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles Al(OH)3 = 30.0 grams / 78.0 g/mol

Moles Al(OH)3 = 0.385 moles

Moles H2SO4 = 35.0 grams / 98.08 g/mol

Moles H2SO4 = 0.357 moles

Step 4: Calculate the limiting reactant

For 2 moles Al(OH)3 we need 3 moles H2SO4 to produce 1 mol Al(SO4)3 and 6 moles H20

H2SO4 is the limiting reactant. It will completely be consumed ( 0.357 moles). Al(OH)3 is in excess. There will be react 2/3 * 0.357 = 0.238 moles

There will remain 0.385 - 0.238 = 0.147 moles

Mass of Al(OH)3 remaining = 0.147 moles* 78.0 g/mol = 11.47 grams

There will remain 11.47 grams of Al(OH)3

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Which of these changes occurs during a nuclear change?

Andrews [41]

The correct answer is D. Elements are created that differ from the reactants.. Because in a eeaction, the product is different than the reactants. Take NaCl for example.

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3 years ago

Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g)

klio [65]

Answer:

$Kp=3.07x10^6$

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

$2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}$

$N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}$

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

$2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2$

Consequently, the equilibrium constant is computed as:

$Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6$

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3 years ago

The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield. How many grams would be prod

Vedmedyk [2.9K]

The grams that would be produced from 7.70 g of butanoic acid and excess ethanol is 7.923grams

calculation

Step 1: write the chemical equation for the reaction

CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH2CH3 +H2O

step 2: find the moles of butanoic acid

moles= mass/ molar mass

= 7.70 g/ 88 g/mol=0.0875 moles

Step 3: use the mole ratio to determine the moles of ethyl butyrate

moles ratio of CH3CH2CH2COOH :CH3CH2CH2COOCH2CH3 is 1:1 therefore the moles of CH3CH2CH2COOCH2CH3 = 0.0875 x78/100=0.0683moles

step 4: find mass = moles x molar mass

= 0.0683 moles x116 g/mol=7.923grams

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3 years ago

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Over the last several decades, scientists have addressed the problem of nonrenewable natural resources such as fossil fuels. Hum

Varvara68 [4.7K]

The answer is D

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3 years ago

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If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount

Schach [20]

<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL

<u>Explanation:</u>

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.16 M

Given mass of manganese (II) nitrate tetrahydrate = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Putting values in above equation, we get:

$0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL$

Volume of water = Volume of solution = 398.4 mL

Hence, the amount of water required to prepare given amount of salt is 398.4 mL

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3 years ago