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Artemon [7]
3 years ago
6

how many grams of calcium sulfate are produced from 10 grams of calcium nitrate and how many grams of calcium sulfate are produc

e from 10 grams of lithium sulfate.
Chemistry
1 answer:
AlekseyPX3 years ago
6 0

Answer:  8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Ca(NO_3)_2=\frac{10g}{164g/mol}=0.061moles

\text{Moles of} Li_2SO_4=\frac{10g}{110g/mol}=0.091moles

Ca(NO_3)_2+Li_2SO_4\rightarrow 2LiNO_3+CaSO_4  

According to stoichiometry :

1 mole of Ca(NO_3)_2 require = 1 mole of Li_2SO_4

Thus 0.061 moles of Ca(NO_3)_2 will require=\frac{1}{1}\times 0.061=0.061moles  of Li_2SO_4

Thus Ca(NO_3)_2 is the limiting reagent as it limits the formation of product and Li_2SO_4 is the excess reagent.

As 1 mole of Ca(NO_3)_2 give = 1 mole of CaSO_4

Thus 0.061 moles of Ca(NO_3)_2 give =\frac{1}{1}\times 0.061=0.061moles  of CaSO_4

Mass of CaSO_4=moles\times {\text {Molar mass}}=0.061moles\times 136g/mol=8.30g

Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

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This green solution of chromium(III) can further be reduced by zinc metal to a blue solution of chromium(II) ions. Write the bal
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Answer:

Half-reactions:

Cr³⁺ + 1e⁻ → Cr²⁺; Zn → Zn²⁺ + 2e⁻

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Explanation:

The Cr³⁺ is reduced to Cr²⁺:

<h3>Cr³⁺ + 1e⁻ → Cr²⁺ -Half-reaction 1-</h3>

Zn is oxidized to Zn²⁺:

<h3>Zn → Zn²⁺ + 2e⁻ -Half-reaction 2-</h3>

Twice the reduction of Cr:

2Cr³⁺ + 2e⁻ → 2Cr²⁺

Now this reaction + Oxidation of Zn:

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Which of the following equations does not demonstrate the law of conservation of mass?
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The third option does not obey the law of conservation of mass.

Option 3.

Explanation:

The law of conservation of mass states that the sum of the masses of reactants should be equal to the sum of the masses of the products.

For example, if we consider the first option to verify if it obeys law of conservation of mass or not, 2 Na + Cl₂ → 2 NaCl

So one way to verify it is to find the mass of Na, then multiply it with 2, and then add this with 2 times of mass of chlorine. So this sum should be equal to the 2 times mass of NaCl. But it is somewhat lengthy.

Another way to easily determine this is to check if the elements are present equally in both sides. Such as, in reactant side and product side 2 atoms of Na is present . Similarly, the Cl atoms are also present in equal number in both reactant and product side. Thus this obeyed the law of conservation of mass.

Like this, if we see the second option, there also 1 atom of Na is present in reactant and product side and 2 molecules of H is present in reactant and product side, 1 oxygen is present in reactant and product side and 1 Cl is present in reactant and product side. So it also obeys the law of conservation of mass.

But in the third option, P₄ + 5 O₂→ 2 P₄O₁₀, here, there is 4 atoms of P in reactant side but in product side there is (4*2) = 8 atoms of P. Similarly, the number of atoms of oxygen in reactants and product side is also not same. So the third option does not obey the law of conservation of mass.

The fourth option also obeys the law of conservation of mass as the number of atoms of each element is same in both the product and reactant side.

Thus, the third option does not obey the law of conservation of mass.

5 0
3 years ago
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