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3 years ago

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Calcium hydroxide reacts with carbon dioxide to form calcium carbonate (balanced equation given below). if you start with 155 ml

of a 1.25 m solution of calcium hydroxide and add 10.0 g of carbon dioxide, how much calcium carbonate would be formed?

1 answer:

Katyanochek1 [597]3 years ago

5 0

I think it is 20.42 g I don't know hope I've helped

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Please help, I'm struggling with stoichiometry

rusak2 [61]

Answer:

a) HNO3

b) 26.8g (3 s.f.)

c) 1.29g (3 s.f.)

Please see the attached pictures for full solution.

To balance an equation, ensure that the number of atoms for each element is the same on both sides.

3 0

3 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

3 years ago

Which statement best explains how the solution should be made?

MissTica

Answer:

B

Explanation:

Correct on Edge

8 0

2 years ago

Read 2 more answers

aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid wat

aleksandr82 [10.1K]

Answer:

Explanation:

HCl + NaOH = NaCl + H₂O.

1 mole 1 mole 1 mole 1 mole

6.93 g of hydrochloric acid = 6.93 / 36.5 = .189 mole of HCl

2.4 g of NaOH = 2.4 / 40 = .06 mole of NaOH

NaOH is in short supply so it is the limiting reagent .

1 mole of NaOH reacts with 1 mole of HCl to give 1 mole of Water

.06 mole of NaOH will react with .06 mole of HCl to give .06 mole of water

Water formed = .06 mole

= .06 x 18 = 1.08 g

= 1.1 g

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2 years ago

WILL MARK BRAINLIEST please

astra-53 [7]

Answer:

C is the only reasonable answer.. but this is 6th grade science and I'm in 7th, so I'm pretty sure I'm right

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3 years ago