All categories

3 years ago

How much CuSO4*5H2O and Na2CO3 is needed to obtain 5g of CuCO3?

1 answer:

5 0

I think the reaction that is occurring is CuSO₄(aq)+Na₂CO₃(aq)⇒CuCO₃(s)+Na₂SO₄(aq). Water in the hydrate will just become part of water in solution so it does not really matter for this question.
SInce we now know the balanced chemical equation, you can use stoichiometry to find the amount of reactants needed to produce 5g CuCO₃.
The first step to any stoichiometry question is to convert the mass given into moles. To do this you have to divide the mass by the molar mass of the compound the mass is referring to. The molar mass of CuCO₃ is 123.5g/mol so you have to divide 5g by 123.5g/mol to get 0.04047 moles of CuCO₃.
The next step in this qustoin is to find the number of moles of each reactant using the moles of product we found in the first step. To do this, we need go back and look at the equation to find the molar ratios which is shown through the coefficients. Since all of the coeficients in the chemical equation is 1, we know the molar ratios are all `1 to 1 and therefore the number of moels of each reactant is equal to the number of moles of the product found in step one. This means that you started off with 0.04047 moles of both CuSO₄ and Na₂CO₃.
The final step is to multiply the number moles of each reactant by its molar mass. the molar mass of CuSO₄ is 159.6g/mol and the molar mass of Na₂CO₃ is 106g/mol. When you multiply 0.04047mol by 159.6g/mol you get 6.4g and when you multiply 0.04047mol by 106g/mol you get 4.29g.
Therefore you started with 6.4g of CuSO₄ and 4.29g of Na₂CO₃.
I hope this helps.

You might be interested in

Explain why the atomic radius of fluorine is smaller than that of lithium

Sladkaya [172]

1. Common knowledge : Go to the right of periodic table, the atomic radius is decreasing

2. Flurine has 9 protons and lithium has 3 protons. you know that the electron is attracted with the centre of the atom, that's why more proton, more 'energy' that attract to the centre and that's why it make the shape of the atom is smaller

4 0

2 years ago

Which boiling point is lower an why?

DiKsa [7]

Answer:4

Explanation:

If we carefully observe the electronegativity of the elements in question

P-2.19

N-3.04

C-2.55

Si-1.9

H-2.2

SiH4 is definitely more polar than CH4 hence greater dipole forces of a higher boiling point. NH3 is more polar than PH3 hence NH3 has greater dipole forces and a higher boiling point. Electronegative differences influences the polarity of a bond. The greater the electro negativity difference between bonding atoms, the greater the dipole forces and the greater the boiling point.

3 0

3 years ago

Read 2 more answers

In the world of chemisty, a 'mole is a number of something ... a very lary large number of something.

yulyashka [42]

Just as a dozen is 12 of something, a mole is about 6.02 * 10^23 of something .

5 0

2 years ago

What type of energy is present in a barbell being lifted or a shot-put being thrown?

denpristay [2]

The type of energy used is kinetic energy. Kinetic energy is the energy of motion.

8 0

3 years ago

Read 2 more answers

I need help with this assignment

yuradex [85]

The answers for the following sums is given below.

1. $Pd_{2}H_{2}$

2. $C_{2} H_{6}$

3. $C_{2} H_{2} O_{2} Cl_{2}$

4. $C_{3}Cl_{3} N_{3}$

5. $Tl_{2 } C_{4} H_{4}O_{6}$

6. $C_{8}H_{8}$

7. $N_{2}O_{5}$

8. $P_{4}O_{6}$

9. $C_{4}H_{8} O_{2}$

Explanation:

1.Given:

Molar mass=216.8g

Molecular formula=Pd $H_{2}$

we know;

Molecular formula=n(Empirical formula)

molecular weight of palladium(Pd)=106.4u

molecular weight of hydrogen(H)=1u

Molar mass of Pd $H_{2}$ :

Pd=106.4×1=106.4u

H=1×2=2

molar mass of Pd $H_{2}$ =106.4+2=108.4

n= $\frac{216.8}{106.4}$

n=2

Molecular formula=2(Pd $H_{2}$ )

Molecular formula= $Pd_{2}H_{2}$

Therefore the molecular formula of the compound is $Pd_{2}H_{2}$

2. Given:

Molar mass=30.0g

Molecular formula= $CH_{3}$

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

Molar mass of $CH_{3}$ :

C=12.01 × 1 = 12.01u

H=1 × 3 = 3u

molar mass of $CH_{3}$ =12.01 + 3 =15.01u

n= $\frac{30.0}{15.01}$

n=2

Molecular formula=2( $CH_{3}$ )

Molecular formula= $C_{2} H_{6}$

Therefore the molecular formula of the compound is $C_{2} H_{6}$

3. Given:

Molar mass=129g

Molecular formula=CHOCl

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

molecular weight of chlorie(Cl)=35.5u

Molar mass of CHOCl:

C=12.01 × 1 = 12.01u

H=1 × 1 = 1u

O=16.00×1=16.00u

Cl=35.5×1=35.5u

molar mass of CHOCl=12.01+1+16.00+35.5=64.5u

n= $\frac{129}{64.5}$

n=2

Molecular formula=2(CHOCl)

Molecular formula= $C_{2} H_{2} O_{2} Cl_{2}$

Therefore the molecular formula of the compound is $C_{2} H_{2} O_{2} Cl_{2}$

5. Given:

Molar mass=577g

Molecular formula= $TlC_{2} H_{2}O_{3}$

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Thallium(Tl)=204.3u

molecular weight of hydrogen(H)=1u

molecular weight of oxygen(O)=16.00u

Molar mass of $TlC_{2} H_{2}O_{3}$ :

C=12.01 × 2= 24.02u

Tl=204.3×1=204.3u

H=1×2=2u

O=16.00×3=48.00

molar mass of $TlC_{2} H_{2}O_{3}$ =204.3+24.02+1+48.00=278.32u

n= $\frac{577}{278.32}$

n=2

Molecular formula=2 ( $TlC_{2} H_{2}O_{3}$ )

Molecular formula= $Tl_{2 } C_{4} H_{4}O_{6}$

Therefore the molecular formula of the compound is $Tl_{2 } C_{4} H_{4}O_{6}$

4. Molar mass=184.5g

Molecular formula=CClN

we know;

Molecular formula = n (Empirical formula)

molecular weight of Carbon(C)=12.01u

molecular weight of Nitrogen(N)=14u

molecular weight of chlorine(Cl)=35.5u

Molar mass of CClN:

C=12.01 × 1 = 12.01u

N=1×14=14U

Cl=35.5×1=35.5u

molar mass of CClN=12.01+14+35.5=61.5u

n= $\frac{184.5}{61.5}$

n=3

Molecular formula=3 (CClN)

Molecular formula= $C_{3}Cl_{3} N_{3}$

Therefore the molecular formula of the compound is $C_{3}Cl_{3} N_{3}$

6. For the table refer the attached file.

Simplest ratio of elements:

Carbon=8

Hydrogen=8

Empirical formula= $C_{8}H_{8}$

Molecular formula = $C_{8}H_{8}$

Molar mass of $C_{8}H_{8}$ :

molecular weight of carbon=12.04u

molecular weight of hydrogen=1u

C=8×12.01=96.08u

H=1×8=8u

molar mass of $C_{8}H_{8}$ =96.08+8=104.08u

n=104.08÷78.0

n=1

Molecular formula = n(Empirical formula)

Molecular formula = 1( $C_{8}H_{8}$ )

Molecular formula = $C_{8}H_{8}$

Therefore the molecular formula of a compound is $C_{8}H_{8}$

7. Given:

mass of oxide of nitrogen=108g

mass of nitrogen=4.02g

mass of oxygen=11.48g

moles of nitrogen= $\frac{4.04}{14.01}$ = 0.289 moles

moles of oxygen= $\frac{11.46}{15.999}$ =0.716 moles

We divide through by the lowest molar quantity to give an empirical formula of $N_{2} O{5}$ .

Now the molecular formula is multiple of the empirical formula.

So,

108 = n × (2×14.01 + 5×15.999)

Clearly,n=1, and the molecular formula is $N_{2}O_{5}$ .

8.For the table refer the attached file.

Simplest ratio of elements:

Phosphorus=2

Oxygen=3

We know;

Empirical formula= $P_{2} O_{3}$

molecular formula= 2(Empirical formula)

Molecular formula =2( $P_{2} O_{3}$ )

Molecular formula = $P_{4}O_{6}$

Therefore the molecular formula of the compound is $P_{4}O_{6}$

9. For the table refer the attached file.

Simplest ratio of elements:

Carbon=2

Hydrogen=9

Oxygen=2

We know;

Empirical formula = $C_{2} H_{4} O$

Molecular formula = 2(Empirical formula)

Molecular formula =2( $C_{2} H_{4} O$ )

Molecular formula = $C_{4}H_{8} O_{2}$

Therefore the molecular formula of the compound is $C_{4}H_{8} O_{2}$

4 0

3 years ago