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mart [117]
11 months ago
11

(d) How many grams of CaO can be produced?

Chemistry
1 answer:
Hunter-Best [27]11 months ago
3 0

218.4 grams of CaO is produced using 3.9 moles CaCO₃.

<h3>How we calculate weight of any substance from moles?</h3>

Moles of any substance will be defined as:

n = W / M

Given chemical reaction is:

CaCO₃ → CaO + CO₂

From the above equation it is clear that according to the concept of stoichiometry 1 mole of CaCO₃ is producing 1 mole of CaO. By using above formula, we calculate grams as follow:

W = n × M, where

n = no. of moles of CaO = 3.9 moles

M = molar mass of CaO = 56 g/mole

W = 3.9 × 56 = 218.4 g

Hence, 218.4 grams of Cao is produced.

<h3>How much grams do Cao's molecular weight equal?</h3>

Molecular weight of CaO. CaO has a molar mass of 56.0774 g/mol. Calcium Oxide is another name for this substance. Convert moles of CaO to grams or grams of CaO to moles. Calculation of the molecular weight: 40.078 + 15.9994 ›› Composition by percentage and element

<h3>How much calcium is required to create one mole of oxygen?</h3>

In order to create one mole of calcium oxide, the reaction between one mole of calcium (40.1 g) and half a mole of oxygen (16 g) is stoichiometric (56.1 g). This means that only 4.01 grams of calcium metal and 1.6 grams of oxygen can combine to generate 5.61 grams of calcium oxide.

Learn more about moles:

brainly.com/question/13314627

#SPJ4

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Answer:

\boxed{\text{47.4 g}}

Explanation:

We are given the mass of two reactants, so this is a limiting reactant problem.

We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:    17.03   32.00                 18.02  

           4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

m/g:     70.1      70.1

Step 1. Calculate the moles of each reactant

\text{Moles of CO } = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{4.116 mol}\\\\\text{Moles of H$_{2}$O} = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{2.191 mol}

Step 2. Identify the limiting reactant  

Calculate the moles of H₂O we can obtain from each reactant.

From NH₃:

The molar ratio of H₂O:NH₃ is 6:4.

\text{Moles of H$_{2}$O} = \text{4.116 mol NH$_{3}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{4 mol NH$_{3}$}} = \text{6.174 mol H$_{2}$O}

From O₂:  

The molar ratio of H₂O:O₂ is 6:5.  

\text{Moles of H$_{2}$O} = \text{2.191 mol O$_{2}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{5 mol O$_{2}$}} = \text{2.629 mol H$_{2}$O}

O₂ is the limiting reactant because it gives the smaller amount of H₂O.  

Step 3. Calculate the theoretical yield.

\text{Theor. yield } = \text{2.629 mol H$_{2}$O}\times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{47.4 g H$_{2}$O}\\\\\text{The maximum yield of H$_{2}$O is }\boxed{\textbf{47.4 g}}

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Answer:

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