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Zigmanuir [339]
3 years ago
8

calculate the number of grams of magnesium in 50g of magnesium sulfide. Use percentage composition and gravimetric factor.

Chemistry
1 answer:
faltersainse [42]3 years ago
3 0

Answer:

50 gram of MgS contains =21.56 gram of Mg

Explanation:

Molecular mass of MgS is 56.37 g/mol.

Atomic mass of 1 mole Mg is 24.305

Therefore 56.37 gram of MgS contains 24.305 gram of Mg

                  1   gram of MgS contains \frac{24.305}{56.37} gram of Mg

                  50 gram of MgS contains  \frac{24.305}{56.37}  ×50 gram of Mg

                                                             =21.56 gram of Mg

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7 0
3 years ago
4HCl(g)+O2(g)⟶2Cl2(g)+2H2O(g) Calculate the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2. mass:
FromTheMoon [43]

The number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Given data:

Moles of hydrochloric acid = 0.385 mol

Mass of chlorine gas =?

Chemical equation:

4HCl +  O₂ → 2Cl₂  + 2H₂O

Now we will compare the moles of Cl₂ with HCl.

                 HCl           :               Cl₂

                   4             :               2

                0.385       :              2÷4× 0.385 = 0.1925 mol

Oxygen is present in excess that's why the mass of chlorine produced depends upon the available amount of HCl.

Mass of Cl₂ :

Mass of Cl₂ = moles × molar mass

Mass of Cl₂ =0.1925 mol × 71 g/mol

Mass of Cl₂ =  13.6675 g

Hence, the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

6 0
2 years ago
Please help
ludmilkaskok [199]

Answer: The molecular formula will be C_3H_9

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.  

The empirical formula is CH_3

The empirical weight of CH_3 = 1(12)+3(1)= 15g.

The molecular weight = 45.0 g/mole

Now we have to calculate the molecular formula:

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{45}{15}=3

The molecular formula will be=3\times CH_3=C_3H_9

Thus molecular formula will be C_3H_9

6 0
3 years ago
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