Chemistry, 24.03.2020 19:34 aero8030

For the following reaction, 9.37 grams of carbon (graphite) are allowed to react with 12.6 grams of oxygen gas . carbon (graphite)(s) + oxygen(g) carbon dioxide(g) What is the maximum mass of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams Submit AnswerRetry Entire Group

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Answer from: Star6147

The maximum mass of carbon dioxide formed is 17.34 grams, the formula of limiting reagent is O_2 and the mass of excess reagent (carbon) remained is 4.644 grams

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For carbon:

Given mass of carbon = 9.37 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon}=\frac{9.37g}{12g/mol}=0.781mol$

For oxygen gas:

Given mass of oxygen gas = 12.6 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{12.6g}{32g/mol}=0.394mol$

The chemical equation for the reaction of carbon and oxygen gas follows:

$C(s)+O_2(g)\rightarrow CO_2(g)$

By Stoichiometry of the reaction:

1 moles of oxygen gas reacts with 1 mole of carbon metal.

So, 0.394 moles of oxygen gas will react with = $\frac{1}{1}\times 0.394=0.394mol$ of carbon metal.

As, given amount of carbon metal is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (carbon) left = [0.781 - 0.394] = 0.387 moles

By Stoichiometry of the reaction:

1 moles of oxygen gas produces 1 mole of carbon dioxide

So, 0.394 moles of oxygen gas will produce = $\frac{1}{1}\times 0.394=0.394moles$ of carbon dioxide

Now, calculating the mass of carbon and carbon dioxide from equation 1, we get:

For carbon:

Excess moles of carbon = 0.387 moles

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

$0.387mol=\frac{\text{Mass of carbon}}{12g/mol}\\\\\text{Mass of carbon}=(0.387mol\times 12g/mol)=4.644g$

For carbon dioxide:

Moles of carbon dioxide = 0.394 moles

Molar mass of carbon dioxide = 44 g/mol

Putting values in equation 1, we get:

$0.394mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.394mol\times 4g/mol)=17.34g$

Hence, the maximum mass of carbon dioxide formed is 17.34 grams, the formula of limiting reagent is O_2 and the mass of excess reagent (carbon) remained is 4.644 grams

Answer from: Quest

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

Answer from: Quest

10 years is the answer

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