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3 years ago

Which of the following coefficients are needed to balance the combustion reaction of ethanol?

1 answer:

5 0

Answer is: d. 3,2,3.

Balanced chemical reaction: CH₃CH₂OH + 3O₂ → 2CO₂ + 3H₂O.
CH₃CH₂OH is ethanol.
O₂ is molecule of oxygen.
CO₂ is carbon(IV) oxide.
H₂O is water.
There are same number of atoms (oxygen, carbon and hydrogen) on both side of balanced chemical reaction:
2 atoms of carbon.
6 atoms of hydrogen.
7 atoms of oxygen.

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Magnesium hydroxide [Mg(OH)2] is an ingredient in some antacids. How many grams of Mg(OH)2 are needed to neutralize the acid in

Allushta [10]

Answer:

1.95g of Mg(OH)2 are needed

Explanation:

Mg(OH)2 reacts with HCl as follows:

Mg(OH)2 + 2 HCl → MgCl2 + 2H2O

<em>Where 1 mole of Mg(OH)2 reacts with 2 moles of HCl</em>

To solve this question we must find the moles of acid. Then, with the chemical equation we can find the moles of Mg(OH)2 and its mass:

<em>Moles HCl:</em>

158mL = 0.158L * (0.106mol / L) = 0.01675 moles HCl

<em>Moles Mg(OH)2:</em>

0.01675 moles HCl * (2mol Mg(OH)2 / 1mol HCl) = 0.3350 moles Mg(OH)2

<em>Mass Mg(OH)2 -Molar mass: 58.3197g/mol-</em>

0.3350 moles Mg(OH)2 * (58.3197g / mol) =

<h3>1.95g of Mg(OH)2 are needed</h3>

3 0

3 years ago

Which models are most appropriate for him to use?

jekas [21]

Answer:

the triangular model and the cubic model

5 0

3 years ago

Read 2 more answers

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$