When a sample of magnesium nitrate was heated, some of the substance decomposed. The mass of the solid residue was 15.4 g.

When a sample of magnesium nitrate was heated, some of the substance decomposed. The mass of the solid residue was 15.4 g. This residue can be reacted with 20 g of 20% sodium hydroxide solution. Determine the mass of the original sample and the volume of evolved gases (in terms of n.u.).

Let us write the equation for the thermal decomposition of magnesium nitrate:
2Mg (NO3) 2 → t 2MgO + 4NO2 + O2 + {Mg (NO3) 2} rest. (1),
where {Cu (NO3) 2} rest. – undecomposed part of magnesium nitrate.
Thus, the solid residue is a mixture of the formed magnesium oxide and the remaining magnesium nitrate. Only one component of the solid residue reacts with sodium hydroxide – the remaining Mg (NO3) 2:
Mg (NO3) 2 + 2NaOH → Mg (OH) 2 + 2NaNO3 (2)
Let’s find the amount of the substance and the mass of sodium hydroxide:
m (NaOH) = m (NaOH) solution ∙ ω (NaOH) = 20 g ∙ 0.2 = 4 g
n (NaOH). = m (NaOH) / M (NaOH) = 4 g / 40 g / mol = 0.1 mol.
From equation (2): n (Mg (NO3) 2) rest. = 0.5 ∙ n (NaOH) = 0.5 ∙ 0.1 mol = 0.05 mol,
m (Mg (NO3) 2) rest. = n (Mg (NO3) 2) rest. ∙ M (Mg (NO3) 2) = 0.05 mol, ∙ 148 g / mol = 7.4 g.
Find the mass and amount of the magnesium oxide substance:
m (MgO) = residual – m (Mg (NO3) 2) resid. = 15.4g – 7.4g = 8g.
n (MgO). = m (MgO) / M (MgO) = 8g / 40 g / mol = 0.2 mol.
Let’s find the amount of substance and the volume of the gas mixture:
From equation (1): n (NO2 + O2) = 5/2 ∙ n (CuO) = 5/2 ∙ 0.2 mol = 0.5 mol.
V (NO2 + O2) = n (NO2 + O2) ∙ Vm = 0.5 mol ∙ 22.4 l / mol = 11.2 l.
Let’s find the amount of the substance and the mass of the original magnesium carbonate:
Equations (1): n (Mg (NO3) 2) reac. = n (MgO) = 0.2 mol.
m (Mg (NO3) 2) reac. = n (Mg (NO3) 2) reac. ∙ M (Mg (NO3) 2) = 0.2 mol, ∙ 148 g / mol = 29.6 g.
m (Mg (NO3) 2) ref. = m (Mg (NO3) 2) reac. + m (Mg (NO3) 2) rest = 29.6 g + 7.4 g = 37g.
Answer: V (NO2 + O2) = 11.2 l; m (Mg (NO3) 2) = 37 g.

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