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3 years ago

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Diborane (B2H6) is a gas at room temperature that forms explosive mixtures with air. It reacts with oxygen according to the foll

owing equation (which may or may not be balanced): _____ B2H6 (g) + _____ O2 (g) → _____ B2O3 (s) + _____ H2O (l) How many grams of O2 (molar mass 32.00 g/mol) will react with 14.67 grams of diborane (molar mass 27.67 g/mol). Your answer must be expressed to the correct number of significant figures, and with the correct unit.

1 answer:

stiv31 [10]3 years ago

3 0

Answer: 50.91 grams

Explanation:

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

$\text{Moles of} B_2H_6=\frac{14.67g}{27.67g/mol}=0.5302moles$

The balanced chemical equation is :

$B_2H_6(g)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

According to stoichiometry :

1 mole of $B_2H_6$ require = 3 moles of $O_2$

Thus 0.5302 moles of $B_2H_6$ will require = $\frac{3}{1}\times 0.5302=1.591moles$ of $O_2$

Mass of $O_2=moles\times {\text {Molar mass}}=1.591moles\times 32.00g/mol=50.91g$

Thus 50.91 grams of $O_2$ reacts with 14.67 g of diborane

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20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc

bezimeni [28]

Answer:

Concentration of the barium ions = $[Ba^{2+}] = 0.4654 M$

Concentration of the chloride ions = $[Cl^{-}]=0.9308 M$

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride = 0.1355 M

$n=0.1355 M\times 0.04389 L=0.005947 mol$

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

$\frac{1}{2}\times 0.05947 mol=0.029735 mol$ barium hydroxide

Moles of barium hydroxide = 0.029735 mol

$Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)$

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

$=1\times 0.029735 mol= 0.029735 mol$

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL = 63.89 mL = 0.06389 L

Concentration of the barium ions = $[Ba^{2+}]$

$[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M$

$Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)$

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

$[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M$

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3 years ago

What process is used to release energy in nuclear power plants?

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Nuclear fission is used enable to release energy in power plants. The constant collision of particles within the reactor, create most of the plants energy.

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3 years ago

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How many moles of KNO3 are in 500 mL of 2. 0 M KNO3? mol KNO3.

8090 [49]

The moles can be defined as the mass of the substance with respect to molar mass. The moles of potassium nitrate is 1 mol.

<h3>How to calculate moles of a substance?</h3>

The moles of a compound can be calculated from:

$\rm Moles=\dfrac{Mass}{Molar\;mass}$

The molarity can be defined as the moles of solute in a liter of solution.

The molarity can be expressed as:

$\rm Molarity=\dfrac{Moles\;\times\;1000}{Volume\;(mL)}$

The molarity of potassium nitrate solution is 2 M, and the volume is 500 mL.

The moles of potassium nitrate is given as:

$\rm 2\;M=\dfrac{Moles\;\times\;1000}{500\;mL}\\ Moles=\dfrac{2\;\times\;500}{1000}\;mol\\ Moles=1\;mol$

The moles of potassium nitrate in 2 M, 500 mL solution are 1 mol.

Learn more about moles, here:

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2 years ago

During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that

bonufazy [111]

Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction: 2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

<em>N₂O₄ is the limiting reactant</em>

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ = 46.15 g of N₂

In the side reaction, (6 * 30 g) of NO is produced from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ = 40.23 g of N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

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2 years ago

How many moles does the mass quantity 130g of H2O have?

SVETLANKA909090 [29]

It has 7.22 moles.... mole =mass/molar mass
molar mass of H2O=18
130/18=mole

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3 years ago