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2 years ago

How many grams of SO2 can be produced if 2.5 moles of O2 are used? S2 + 2O2 → 2SO2

Chemistry

1 answer:

loris [4]2 years ago

6 0

Answer: mass of SO2 is 160 g

Explanation: according to equation number of moles of SO2

Is same as amount of Oxygen, also 2.5 moles.

Molar mass of SO2 is 32.07 +2·16 = 64.07 g/mol

Mass is m= nM = 2.5 mol· 64.07 g/mol= 160.175 g

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How many grams are there in 7.3 moles of NH3?

Schach [20]

Answer:

124.322796 grams

Explanation:

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3 years ago

What is an ionic solution and how does it work in a battery?

frutty [35]

Explanation:

An ionic solution is a solution that has anions and cations. This is formed when ionic compounds dissociate in water. The ionic solution can be used as a battery because their respective ions can be attracted to either node of a battery. The cathode (+) attracts anions (negatively charged ions) and the anode (-) attracted cations (positively-charged ions). The reaction between the nodes and the respective ions in the solution causes electrons do be either donated or received. This ensures that the ionic solution (electrolyte) acts as channels for transport of electrons across the nodes to complete a circuit.

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3 years ago

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 $NaCl$