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Wittaler [7]
3 years ago
5

. Which substance is the limiting reactant when 2 g of S reacts with 3 g of O2 and 4 g of NaOH according to the following chemic

al equation: 2 S (s) + 3 O2 (g) + 4 NaOH (aq) → 2 Na2SO4 (aq) + 2 H2O (l) a) S (s) b) O2 (g) c) NaOH (aq) d) none of these substances is the limiting reactant.
Chemistry
1 answer:
Ksivusya [100]3 years ago
3 0

Answer:

S is the limiting reagent.

Explanation:

To find the limiting reactant we must first write the balanced chemical reaction. It must be correctly balanced so that we can find the proper mole ratios.

2 S (s) + 3 O2 (g) + 4 NaOH (aq) → 2 Na2SO4 (aq) + 2 H2O (l)

After this we will convert our measurements to moles. For mass we do this by dividing by the molar mass.

2g ÷ 32.06 = 0.06238mol S

3g ÷ 32.00 = 0.09375mol O₂

Now that we have the moles of each of the reactants, we can multiply them by their mole ratio with a reactant.

0.06238mol S × 2/2 = 0.06238mol H2O

0.09375mol O₂ × 2/3 = 0.06250mol H2O

S is our limiting reagent because it makes the smaller amount of moles.

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Na2SO4<br> How many oxygen stems are?
IgorLugansk [536]

Answer:

Na2SO4 means: two moles sodium (45.98 g), one mole sulfur (32.06 g), and four moles oxygen (64.00 g) combine to form one mole of sodium sulfate (142.04 g).

Explanation:

3 0
3 years ago
Read 2 more answers
Boron carbonate decomposes into boron oxide and carbon dioxide. How many moles of CO2
Paul [167]
<h3>Answer:</h3>

15 moles

<h3>Explanation:</h3>

The decomposition of boron carbonate is given by the equation;

B₂(CO₃)₃(s) → B₂O₃(s) + 3CO₂(g)

Moles of boron carbonate decomposed is 5.0 mol

To find the moles of CO₂ produced we are going to use the mole ratio.

Mole ratio of B₂(CO₃)₃ to CO₂ is 1 : 3

Therefore;

Moles of CO₂ = Moles of B₂(CO₃)₃ × 3

                      = 15 mol

Therefore, 15 moles of CO₂ will be produced

   

4 0
3 years ago
10. Which term is same for one mole of O2 and one mole of ethane C2H6
melamori03 [73]

Answer:

b) Number of molecules

Explanation:

3 0
3 years ago
Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3?
Helen [10]

Answer:

41 mL

Explanation:

Given data:

Milliliter of HCl required = ?

Molarity of HCl solution = 4.25 M

Mass of CaCO₃ = 8.75 g

Solution:

Chemical equation:

2HCl + CaCO₃      →    CaCl₂ + CO₂ + H₂O

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 8.75 g / 100.1 g/mol

Number of moles = 0.087 g /mol

Now we will compare the moles of  CaCO₃ with HCl.

                      CaCO₃         :          HCl

                          1               :            2

                      0.087           :         2/1×0.087 = 0.174 mol

Volume of HCl:

Molarity = number of moles / volume in L

4.25 M = 0.174 mol / volume in L

Volume in L = 0.174 mol /4.25 M

Volume in L = 0.041 L

Volume in mL:

0.041 L×1000 mL/ 1L

41 mL

8 0
3 years ago
In an chemicalbreaction involving Fe and S, it was found that 45.2 g of Fes was produced. If the percent yield of the reaction i
saw5 [17]

Answer:

47.8 g

Explanation:

Remember the equation for percent yield:

% yield = actual / theoretical

We're given two of the values in the question, so plug n' play:

0.945 = 45.2 / theoretical

theoretical = 47.8 g

Keep in mind you can use mass here without converting to moles because we're working with products only. If you were given a mass of reactants, you would need to convert to moles and using a balanced chemical equation find the corresponding moles of product produced.

7 0
3 years ago
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