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Yuki888 [10]
3 years ago
15

Which substance has a molar mass of 142 g?

Chemistry
2 answers:
Nata [24]3 years ago
8 0

Answer:

Molar mass of P₂O₅ = 141.948 g/mol which is approximately 142 g/mol.

Explanation:

Molar mass of AlCl₃

Molar mass  = 26.982 + 35.5 × 3 = 133.32 g/mol

Molar mass of BaCl₂

Molar mass  = 137.33 + 35.5 × 2 = 208.33 g/mol

Molar mass of MgCl₂

Molar mass  = 24.305 + 35.5×2 = 95.305 g/mol

Molar mass of P₂O₅

Molar mass  = 30.974×2 + 16× 5 = 141.948 g/mol

In-s [12.5K]3 years ago
6 0

Answer:

P_2O_5

Explanation:

Molecular formulas is the actual number of atoms of each element in the compound. Molecular mass is the sum of the mass of each element which is present in the molecular formula is multiplied by their subscript which appear in the molecular formula.

The formula is = AlCl_3

Mass from the formula = 1×Mass of Aluminum + Mass of Chlorine×3 = 1×27 + 35.5×3 = 133.5 g/mol

The formula is = BaCl_2

Mass from the formula = 1×Mass of Barium + Mass of Chlorine×2 = 1×137 + 35.5×2 = 208 g/mol

The formula is = MgCl_2

Mass from the formula = 1×Mass of Magnesium + Mass of Chlorine×2 = 1×24 + 35.5×2 = 95 g/mol

The formula is = P_2O_5

Mass from the formula = 2×Mass of Phosphorus + Mass of oxygen×5 = 2×31 + 16×5 = 142 g/mol

<u>Hence, correct answer is:- P_2O_5 </u>

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Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO
sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂   → CaCO₃

Number of moles of CaO:

Number of moles  = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles  = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

                  CO₂         :                CaCO₃  

                  1               :                 1

                 0.31           :              0.31

                CaO           :               CaCO₃  

                 1                :                 1

                 0.26         :              0.26

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ =  26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of  CaCO₃ produced by CaO are less it will be limiting reactant.

7 0
3 years ago
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