Question 7 of 25

2 H2(g) + O2(g) → 2 H2O(g)

alexgriva [62]

Answer:

77 L of water can be made.

Explanation:

Molar mass of $O_{2}$ = 32 g/mol

So, 55 g of $O_{2}$ = $\frac{55}{32}$ mol of $O_{2}$ = 1.72 mol of $O_{2}$

As hydrogen is present in excess amount therefore $O_{2}$ is the limiting reagent.

According to balanced equation, 1 mol of $O_{2}$ produces 2 mol of $H_{2}O$ .

So, 1.72 mol of $O_{2}$ produce $(2\times 1.72)$ mol of $H_{2}O$ or 3.44 mol of $H_{2}O$ .

Let's assume $H_{2}O$ gas behaves ideally at STP.

Then, $P_{H_{2}O}.V_{H_{2}O}=n_{H_{2}O}.R.T$ , where P, V, n, R and T represents pressure, volume, no. of moles, gas constant and temperature in kelvin scale respectively.

At STP, pressure is 1 atm and T is 273 K.

Here, $n_{H_{2}O}$ = 3.44 mol and R = 0.0821 L.atm/(mol.K)

So, $(1atm)\times V_{H_{2}O}=(3.44mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)$

$\Rightarrow$ $V_{H_{2}O}=77L$

Option (b) is correct.

5 0

3 years ago

How much positive charge is in 1.4 kg of oxygen? The atomic weight (15.9994 g) of oxygen contains Avogadro’s number of atoms, wi

Cerrena [4.2K]

Answer:

$6.7511\times 10^7\ C$

Explanation:

The atomic weight of oxygen = 15.9994 g

This mass corresponds to 1 mole of the oxygen atoms.

Thus,

15.9994 g mass of oxygen contains $6.02\times 10^{23}$ atoms of oxygen.

1.4 kg = 1400 g ( 1 kg = 1000 g)

So,

1400 g mass of oxygen contains $\frac {6.02\times 10^{23}}{15.9994}\times 1400$ atoms of oxygen.

Number of atoms in 1400 g of oxygen = $526.769754\times 10^{23}$

Also, 1 atom of oxygen contains 8 protons

Charge of 1 proton = + $1.602\times 10^{-19}\ C$

So, Charge on 1 atom of oxygen = $8\times 1.602\times 10^{-19}\ C$

Thus,

Charge on $526.769754\times 10^{23}$ atoms of oxygen = $526.94476\times 10^{23}\times 8\times 1.602\times 10^{-19}\ C=6.7533\times 10^7\ C$

Thus, positive charge in 1.4 kg of oxygen = $6.7511\times 10^7\ C$

5 0

3 years ago

What mass of silver chloride can be produced from 1.65 l of a 0.240 m solution of silver nitrate?

Liono4ka [1.6K]

Check table T and use the concentration equation. Molarity= moles of solute/ liters of solution.

So 0.240 = x/ 1.65 once u find the # of miles of solute (x=.396) and substitute that

Wait I'm not sure if it's correct

4 0

4 years ago

Suppose 650 mL of hydrogen gas are produced through a displacement reaction involving solid iron and sulfuric acid, H2SO4, at ST

scoray [572]

Answer:

4.415 g of FeSO₄

Explanation:

The balance chemical equation for given single replacement reaction is as follow;

Fe + H₂SO₄ → FeSO₄ + H₂

Data Given;

Volume = 650 mL = 0.65 L

Density at STP = 0.08988 g/L

Mass = Density × Volume = 0.08988 g/L × 0.65 L = 0.0584 g

Step 1: Calculate Moles of H₂ as;

Moles = Mass / M.Mass

Moles = 0.0584 g / 2.01 g/mol

Moles = 0.0290 mol of H₂

Step 2: Find out moles of FeSO₄ as;

According to balance chemical equation,

1 mole of H₂ is produced along with = 1 mole of FeSO₄

So,

0.0290 moles of H₂ will be produced along with = X moles of FeSO₄

Solving for X,

X = 0.0290 × 1 mol / 1 mol

X = 0.0290 moles of FeSO₄

Step 3: Calculate mass of FeSO₄ as;

Mass = Moles × M.Mass

Mass = 0.0290 mol × 151.90 g/mol

Mass = 4.415 g of FeSO₄

4 0

3 years ago

Determine the percent yield for the reaction

kkurt [141]

Answer:

The percent yield for Br₂ in the reaction = 96.15%

Explanation:

The balanced stoichiometric equation for the reaction is given as

2 NaBr + 1 Cl₂ → 2 NaCl + 1 Br₂

Percent yield

= 100% × (Actual yield)/(Theoretical yield)

To find the theoretical yield,

5.29 g of NaBr reacts with excess chlorine

gas; this means that NaBr is the limiting reagent because it is used up in the process of the reaction, hence, it determines the amount of products to be found.

So, we convert the 5.29 g of NaBr into number of moles.

Number of moles = (Mass)/(Molar mass)

Molar Mass of NaBr = 102.894 g/mol

Number of moles = (5.29/102.894) = 0.0514121329 = 0.05141 mole

From the stoichiometric balance of the reaction,

2 moles of NaBr give 1 mole of Br₂

0.05141 mole of NaBr will give (0.05141×1/2) mole of Br₂; that is, 0.0257 mole of Br₂

Theoretical yield = Mass of Br₂ expected from the reaction

= (Number of moles) × (Molar mass)

Molar mass of Br₂ = 159.808 g/mol

Theoretical yield of Br₂ = 0.0257 × 159.808 = 4.108 g

Percent yield

= 100% × (Actual yield)/(Theoretical yield)

Actual yield = 3.95 g

Theoretical yield = 4.108 g

Percent yield = 100% × (3.95/4.108) = 96.15%

Hope this Helps!!!

5 0

4 years ago