Answer
$Ni$ is not the identity of the metal "M", because that would give 0.392 g of $CO_2$ and not 0.376 g.
Work Step by Step
1. Calculate the number of moles of $NiCO_3$:
58.69* 1 + 12.01* 1 + 16* 3 = 118.7g/mol
$1.056g \times \frac{1 mol}{ 118.7g} = 8.896 \times 10^{-3}mol (NiCO_3)$
- The balanced reaction is:
$NiCO_3 -- \gt NiO + CO_2$
The ratio of $NiCO_3$ to $CO_2$ is 1 to 1:
$8.896 \times 10^{-3} mol (NiCO_3) \times \frac{ 1 mol(CO_2)}{ 1 mol (NiCO_3)} = 8.896 \times 10^{-3}mol (CO_2)$
2. Calculate the mass of $CO_2$:
12.01* 1 + 16.00* 2 = 44.01g/mol
$8.896 \times 10^{-3} mol \times \frac{ 44.01 g}{ 1 mol} = 0.392 g (CO_2)$
