#### Answer

3.145 g of Calcium

#### Work Step by Step

1. Write the unbalanced equation.
- Calcium (Ca) reacts with carbon (C) to produce calcium carbide: $(CaC_2)$
$$Ca + C \longrightarrow CaC_2$$
2. Balance the equation.
$$Ca+2C\longrightarrow CaC_2$$
- We just need to balance the number of carbon atoms, by putting 2 as the coefficient for $C$ in the reactants.
3. $ C $ : 12.01 g/mol
$$ \frac{1 \space mol \space C }{ 12.01 \space g \space C } \space and \space \frac{ 12.01 \space g \space C }{1 \space mol \space C }$$
$ Ca $ : 40.08 g/mol
$$ \frac{1 \space mol \space Ca }{ 40.08 \space g \space Ca } \space and \space \frac{ 40.08 \space g \space Ca }{1 \space mol \space Ca }$$
$$ 1.885 \space g \space C \times \frac{1 \space mol \space C }{ 12.01 \space g \space C } \times \frac{ 1 \space mol \space Ca }{ 2 \space mol \space C } \times \frac{ 40.08 \space g \space Ca }{1 \space mol \space Ca } = 3.145 \space g \space Ca $$