Answer
$Ba$ is not the correct identity of the metal, because it would not produce 0.376 g of carbon dioxide.
Work Step by Step
1. Calculate the number of moles of $BaCO_3$:
137.3* 1 + 12.01* 1 + 16* 3 = 197.31g/mol
$1.056g \times \frac{1 mol}{ 197.31g} = 5.352 \times 10^{-3}mol (BaCO_3)$
- The balanced reaction is:
$BaCO_3 -- \gt BaO + CO_2$
The ratio of $BaCO_3$ to $CO_2$ is 1 to 1:
$5.352 \times 10^{-3} mol (BaCO_3) \times \frac{ 1 mol(CO_2)}{ 1 mol (BaCO_3)} = 5.352 \times 10^{-3}mol (CO_2)$
2. Calculate the mass of $CO_2$:
12.01* 1 + 16* 2 = 44.01g/mol
$5.352 \times 10^{-3} mol \times \frac{ 44.01 g}{ 1 mol} = 0.236g (CO_2)$
