Question

when 2.00 g of solid lithium completely reacted with oxygen gas, 4.31 g of solid lithium oxide was produced.

the mass of oxygen gas that reacted in the reaction described was g.

Explanation:

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"Reactants and Products"
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<reasoning>

Identify the reactants and products

\[
\text{Lithium (solid)} + \text{Oxygen (gas)}
ightarrow \text{Lithium oxide (solid)}
\]

Apply the law of conservation of mass

\[
m_{\text{reactants}} = m_{\text{products}}
\]
\[
m_{\text{lithium}} + m_{\text{oxygen}} = m_{\text{lithium oxide}}
\]

Calculate the mass of oxygen gas

\[
2.00\text{ g} + m_{\text{oxygen}} = 4.31\text{ g}
\]
\[
m_{\text{oxygen}} = 4.31\text{ g} - 2.00\text{ g} = 2.31\text{ g}
\]
</reasoning>

<answer>
When \(2.00\text{ g}\) of solid lithium completely reacted with oxygen gas, \(4.31\text{ g}\) of solid lithium oxide was produced.

The mass of oxygen gas that reacted in the reaction described was <blank>2.31</blank> g.
</answer>

<post_analysis>
{
"subject": "Natural Science",
"question_type": "Fill-in-the-blank",
"knowledge_point": [
"Natural Science",
"Chemistry",
"Law of Conservation of Mass"
]
}
</post_analysis>

Answer:

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</pre_analysis>

<reasoning>

Identify the reactants and products

\[
\text{Lithium (solid)} + \text{Oxygen (gas)}
ightarrow \text{Lithium oxide (solid)}
\]

Apply the law of conservation of mass

\[
m_{\text{reactants}} = m_{\text{products}}
\]
\[
m_{\text{lithium}} + m_{\text{oxygen}} = m_{\text{lithium oxide}}
\]

Calculate the mass of oxygen gas

\[
2.00\text{ g} + m_{\text{oxygen}} = 4.31\text{ g}
\]
\[
m_{\text{oxygen}} = 4.31\text{ g} - 2.00\text{ g} = 2.31\text{ g}
\]
</reasoning>

<answer>
When \(2.00\text{ g}\) of solid lithium completely reacted with oxygen gas, \(4.31\text{ g}\) of solid lithium oxide was produced.

The mass of oxygen gas that reacted in the reaction described was <blank>2.31</blank> g.
</answer>

<post_analysis>
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"question_type": "Fill-in-the-blank",
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"Natural Science",
"Chemistry",
"Law of Conservation of Mass"
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