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:

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

$$ 2.00\text{ g} + m_{\text{oxygen}} = 4.31\text{ g} $$

$$ m_{\text{oxygen}} = 4.31\text{ g} - 2.00\text{ g} = 2.31\text{ g} $$

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.