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4 years ago

2 h2 o2---> 2 h2o how many moles of water are produced when 2 moles of hydrogen gas are mixed with 2 moles of oxygen gas?

1 answer:

4 0

2 moles of water are produced. Here, hydrogen is the limiting reagent. 1 mole of oxygen is used up with 2 moles of hydrogen to form 2 moles of water. 1 moles of oxygen remains.

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An electric kettle draws a current of 6.50 A while it is plugged into a 120-V

Umnica [9.8K]

Answer:

780 watts

Explanation:

formula to find power when given amps and voltage

P = A x V

=6.50a x 120V

= 780 W (watts)

7 0

3 years ago

A mixture of 5L of H2 and 3L O2 reacts to form H2O (g) at constant T and P .Find the

slamgirl [31]

The volume of H₂O = 5 L

<h3>Further explanation</h3>

Given

5L of H₂ and 3L O₂

Reaction

2H₂ (g) + O₂(g) ⇒2H₂O(g)

Required

The volume of H₂O

Solution

Avogadro's hypothesis:

<em>In the same T,P and V, the gas contains the same number of molecules </em>

So the ratio of gas volume will be equal to the ratio of gas moles

mol H₂ = 5, mol O₂ = 3

Find limiting reactants

From equation, mol ratio H₂ : O₂ = 2 : 1, so :

$\tt H_2\div O_2=\dfrac{mol~H_2}{coefficient}\div \dfrac{mol~O_2}{coefficient}\\\\=\dfrac{5}{2}\div \dfrac{3}{1}=2.5\div 3\rightarrow H_2~limiting~reactant(smaller~ratio)$

Find volume H₂O

mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L

4 0

3 years ago

How many moles are 4.20 * 10 ^ 25 atoms of Ca?

aleksandr82 [10.1K]

Answer:

~69.744 moles of Ca

Explanation:

Using Avogadro's constant , we know that:

1 mole = 6.022 x 10^23 atoms

S0, the number of moles in 4.20 x 10^25 atoms of Ca:

=(4.20 x 10^25 x 1 )/(6.022 x 10^23)

~69.744 moles of Ca

Q2:How many atoms are in 0.35 moles of oxygen?

1 mole = 6.022 x 10^23 atoms

S0, the number of atoms in 0.35 moles of oxygen:

=[0.35 x (6.022 x 10^23)]

=2.1077 x 10^23 atoms of Oxygen

Hope it helps:)

4 0

3 years ago

Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165 g sample is combusted to produce

mote1985 [20]

Answer:

The empirical formula is C₄H₈O;

The molecular formula is C₈H₁₆O₂.

Explanation:

The empiric formula is CxHyOz, and the combustion reaction occurs between the fuel and oxygen gas, so it will be:

CxHyOz + O₂ → CO₂ + H₂O

So, all the carbon will form CO₂ and all hydrogen will form H₂O. So, let's calculated the number of moles of carbon in CO₂ and the number of moles of hydrogen in H₂O. They must be the same number f the moles in the valproic acid.

The molar masses are: C= 12 g/mol, O = 156 g/mol and H = 1 g/mol, so CO₂ = 12 + 2x16 = 44 g/mol, and H₂O = 2x1 + 16 = 18 g/mol.

nCO₂ = 0.403/44 = 9.16x10⁻³ mol

In 1 mol of CO₂ there is 1 mol of C, so nC = 9.16x10⁻³ mol.

nH₂O = 0.166/18 = 9.22x10⁻³ mol.

In 1 mol of H₂O there are 2 moles of H, so nH = 0.018 mol.

The mass of Carbon and the mass of H in the compound must be:

mC = nxM = 9.22x10⁻³ x12 = 0.1106 g

mH = nxM = 0.018x1 = 0.018 g

The mass of oxygen must be then:

mC + mH +mO = 0.165

mO = 0.165 - 0.1106 - 0.018

mO = 0.0364g

And its number of moles:

nO = 0.0364/16 = 2.275x10⁻³ mol

To have the empirical formula, the coefficients must be the smallest, so let's divide the number of moles for the small one: 2.275x10⁻³

nC = (9.16x10⁻³ )/(2.275x10⁻³) = 4 mol

nH = 0.018/(2.275x10⁻³) = 8 mol

nO = (2.275x10⁻³)/(2.275x10⁻³) = 1 mol

The empirical formula is C₄H₈O.

The molecular formula must be a multiple of the empirical, so it will be n(C₄H₈O). Knowing the molar mass, we can calculate n:

4xnx12 + 8xnx1 + 1xnx16 = 144

48n + 8n + 16n = 144

72n = 144

n = 2

The molecular formula is 2(C₄H₈O) = C₈H₁₆O₂.

3 0

3 years ago

FeCl2(aq) + Na2CO3(aq) FeCO3(s) + 2NaCl(aq) What are the spectator ions in this equation?

irakobra [83]

Answer:

Chloride (Cl⁻) and sodium (Na⁺) ions.

Explanation:

Hello,

In this case, since the aqueous species are actually dissociated when reacting and the solid species (ferric carbonate) remains undissolved, we can modify the given reaction as follows:

$FeCl_2(aq) + Na_2CO_3(aq) \rightarrow FeCO_3(s) + 2NaCl(aq)$

In such a way, dissociating the aqueous species we obtain:

$Fe^{2+}(aq)+2Cl^-(aq) + 2Na^+(aq)+CO_3^{2-}(aq) \rightarrow FeCO_3(s) + 2Na^+(aq)+2Cl^-(aq)$

It means that the net ionic equation is:

$Fe^{2+}(aq)+CO_3^{2-}(aq) \rightarrow FeCO_3(s)$

Therefore, the spectator ions are those were cancelled out, chloride (Cl⁻) and sodium (Na⁺).

Best regards.

8 0

4 years ago