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adelina 88 [10]

3 years ago

9

A chemical equation can be used to solve stoichiometry problems provided that it

2 answers:

Evgesh-ka [11]3 years ago

7 0

Answer: Option (C) is the correct answer.

Explanation:

Stoichiometric coefficients represents the number of individual atoms involved in the chemical equation.

As a result, stoichiometric coefficients help in balancing the equation because an equation can only become balanced when number of reactant molecules equal the number of product molecules or vice versa.

Therefore, we can conclude that a chemical equation can be used to solve stoichiometry problems provided that it includes the coefficients that are needed for balancing the equation.

antoniya [11.8K]3 years ago

6 0

The answer is c. The only things that you need to know is molar ratios which is shown by the coefficients. Everything else only matters when it comes to things like equilibrium.

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The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO)4 (l) → Ni (s) + 4CO (g

Yuki888 [10]

<u>Answer:</u> The volume of CO formed is 254.43 L.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Ni(CO)_4$ = 444 g

Molar mass of $Ni(CO)_4$ = 170.73 g/mol

Putting values in above equation, we get:

$\text{Moles of }Ni(CO)_4=\frac{444g}{170.73g/mol}=2.60mol$

For the given chemical reaction:

$Ni(CO)_4(l)\rightarrow Ni(s)+4CO(g)$

By stoichiometry of the reaction:

1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.

So, 2.60 moles of nickel tetracarbonyl will produce = $\frac{4}{1}\times 2.60=10.4mol$ of carbon monoxide.

Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of the gas = 752 torr

V = Volume of the gas = ? L

n = Number of moles of gas = 10.4 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = Temperature of the gas = $22^oC=(273+22)K=295K$

Putting values in above equation, we get:

$752torr\times V=10.4mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 295K\\\\V=254.43L$

Hence, the volume of CO formed is 254.43 L.

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