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3 years ago

How many grams of MnO, are needed to prepare 10.0 liters of Cl2, gas at 22 C and 710 torr according to the following reaction?

Chemistry

1 answer:

Dominik [7]3 years ago

3 0

Answer:

15.5 g of MnO

Explanation:

The reaction is:

MnO₂ + 2NaCl + 2H₂SO₄ → Cl₂ + MnSO₄ + Na₂SO₄ + 2H₂O

We use the Ideal Gases Law to determine the moles of produced chlorine

P . V = n . R .T

We convert T° to Abs. value → 22°C + 273 = 295K

We convert Torr to atm → 710 Torr . 1atm /760 Torr = 0.934 atm

We replace: 10 L . 0.934 atm = n . 0.082 . 295K

( 10 L . 0.934 atm ) / (0.082 . 295K) = n → 0.386 moles

As the NaCl and the acid are in excess, we assume the MnO is the limiting reactant, and ratio is 1:1. so:

0.386 moles of chlorine came from 0.386 moles of MnO

We convert moles to mass: 0.386 . 40.3g/mol = 15.5 g

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3 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

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3 years ago

37.5g of nitrogen reacts with 15.5 g of hydrogen. What mass of ammonia can be made? What is the limiting reactant?

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Answer:

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3 years ago

The diagram below shows changes to the nucleus of an atom.

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Answer:

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Explanation:

i didn't study it yet sorry for not helping but try asking someone else

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3 years ago

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