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2 years ago

2. Barium compounds are chemically very similar to strontium compounds. Water-soluble

2 answers:

6 0

The ability of sodium sulfate to act as an antidote to barium chloride poisoning is because displaces barium in barium chloride to form sodium chloride and insoluble harmless barium sulfate.

<h3>Why is a solution of sodium sulfate used as an antidote?</h3>

The ability of sodium sulfate to act as an antidote to barium chloride poisoning is because of the reactivity of sodium.

Sodium is a more reactive metal than barium, so it displaces barium in barium chloride to form sodium chloride and insoluble harmless barium sulfate.

The equation of the reaction is given below:

BaCl2 + Na2SO4 ----> BaSO4 + NaCl

Therefore, the ability of sodium sulfate to act as an antidote to barium chloride poisoning is because displaces barium in barium chloride to form sodium chloride and insoluble harmless barium sulfate.

Learn more about about reactivity of metals at: brainly.com/question/24866635

katovenus [111]2 years ago

4 0

There are different kinds of poisonings. Sodium sulfate are known to be an effective antidote because they create or produce a kind of precipitate with barium that is said to be not easily absorbed from the gastrointestinal tract.

There have been a lot of Cases of barium poisoning that are known to be followed by high hypokalemia.

Soluble sulfates such as sodium sulfate and others such as potassium infusion are known to be a key antidote.

If given promptly and in the right dosage, soluble sulfates are known to be an effective antidote.

Learn more about barium from

brainly.com/question/13656770

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How many moles of Ammonium Sulfate can be made from 30.0mol of NH3? 2NH3 + H2SO4 --> (NH4)2SO4

BartSMP [9]

Answer: 15.0 moles of $(NH_4)_2SO_4$ are formed from 30.0 mol of $NH_3$

Explanation:

The balanced chemical reaction is :

$2NH_3+H_2SO_4\rightarrow (NH_4)_2SO_4$

According to stoichiometry :

2 moles of $NH_3$ give = 1 mole of $(NH_4)_2SO_4$

Thus 30.0 moles of $NH_3$ will give = $\frac{1}{2}\times 30.0=15.0moles$ of $(NH_4)_2SO_4$

Thus 15.0 moles of $(NH_4)_2SO_4$ are formed from 30.0 mol of $NH_3$

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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

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