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2 years ago

Al2(SO4)3+Mg(NO3)2⟶ What would be the product(s) of this reaction? These are NOT balanced, just look for the correct products

Chemistry

1 answer:

Readme [11.4K]2 years ago

7 0

Answer:

$Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4$

Explanation:

Hello there!

In this case, for the given reactants side, we infer this is a double replacement reaction because all the cations and anions are switched around as a result of the chemical change, we infer that the products side include aluminum with nitrate and magnesium with sulfate as shown below:

$Al_2(SO_4)_3+Mg(NO_3)_2\rightarrow Al(NO_3)_3+MgSO_4$

However, we need to balance since unequal number of atoms are present at both sides, thus, we do that as shown below:

$Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4$

Thus, we make 6 Al atoms, 3 S atoms, 3 Mg atoms and 30 O atoms on each side in agreement with the law of conservation of mass.

Regards!

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3 years ago

13. A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the

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Answer:

13. 2.60 L.

14. 2.40 L.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and T are constant, and have different values of P and V:

(P₁V₁) = (P₂V₂)

13. A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the pressure is increased to 1.25 atm.

P₁ = 0.755 atm, V₁ = 4.31 L.

P₂= 1.25 atm, V₂ = ??? L.

∴ V₂ = (P₁V₁)/(P₂) = (0.755 atm)(4.31 L)/(1.25 atm) = 2.60 L.

14. 600.0 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at 2.00 at m.

P₁ = 8.0 atm, V₁ = 600.0 mL.

P₂= 2.0 atm, V₂ = ??? L.

∴ V₂ = (P₁V₁)/(P₂) = (8.0 atm)(600.0 mL)/(2.0 atm) = 2400/0 mL = 2.40 L.

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2 years ago

Does a high melting point mean the substance will melt faster or slower?

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2 years ago

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Based on the periodic table why are be, bg, ca and sr in the same column/group/family?

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Answer:

The elements are in the same column/group IIA.

See the explanation below, please.

Explanation:

The elements Calcium, Strontium, Beryllium, Magnesium, Barium and Radio, belong to the group of alkaline earth metals located in group IIA of the periodic table, they require 2 electrons to complete their octet (they have 2 valence electrons). reagents than alkali metals.

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3 years ago

When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain

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Answer:

For a: The number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b: The mass of nitric acid formed is 54.81 grams

For c: The mass of nitric acid formed is 206 grams

Explanation:

The given chemical reaction follows:

$3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)$

For a:

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with $\frac{3}{1}\times 0.250=0.75mol$ of nitrogen dioxide

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.75 moles of nitrogen dioxide will contain $0.75\times 6.022\times 10^{23}=4.52\times 10^{23}$ number of molecules

Hence, the number of molecules of nitrogen dioxide is $4.52\times 10^{23}$

For b:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = $\frac{2}{3}\times 1.304=0.870$ moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

$0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g$

Hence, the mass of nitric acid formed is 54.81 grams

For c:

For nitrogen dioxide:

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol$

For water:

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol$

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = $\frac{1}{3}\times 4.90=1.63mol$ of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce $\frac{2}{3}\times 4.90=3.27mol$ of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

$3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g$

Hence, the mass of nitric acid formed is 206 grams

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3 years ago

Al2(SO4)3+Mg(NO3)2⟶ What would be the product(s) of this reaction? *These are NOT balanced, just look for the correct products*

Al2(SO4)3+Mg(NO3)2⟶ What would be the product(s) of this reaction? These are NOT balanced, just look for the correct products