The mole and the Avogadro constant

Use the relative formula mass of the actual species, not the $A_r$ of one atom. Oxygen gas is $\text{O}_2$, so $M_r = 32$, not 16; hydrogen is $\text{H}_2$, $M_r = 2$; nitrogen is $\text{N}_2$, $M_r = 28$; carbon dioxide is $\text{CO}_2$, $M_r = 44$, not 28. Count every subscript - $\text{Na}_2\text{CO}_3$ is 106, not 83 or 90. A halved $M_r$ doubles every answer after it.

Elemental oxygen, nitrogen and hydrogen exist as diatomic molecules, not lone atoms. Candidates gave N or 2N as the product of reducing NO instead of $\text{N}_2$. The same slip is what drives the molar-mass error: write O rather than $\text{O}_2$ and every mole, mass and gas-volume answer downstream is wrong. Write the formula out in full before you count atoms.

The upside-down formula is one of the digest cross-cutting failures, and it bites hardest here. Amount in moles = mass divided by molar mass. For 50 g of $\text{CaCO}_3$ ($M_r = 100$) that is $50/100 = 0.5$ mol; inverting gives 2 mol - a factor of four out, and every mass or volume derived from it is then wrong. Write the formula down, substitute, and only then divide.

The molar gas volume is 24 dm^3 = 24000 cm^3 per mole - use one of them, not both. Multiplying by 24000 and then by 1000 again is the classic slip. The direction is dm^3 x 1000 = cm^3, and cm^3 divided by 1000 = dm^3, so 6 dm^3 is 6000 cm^3. Then check which unit the question actually asked for: a right number in the wrong unit scores nothing.

The limiting reactant is not the one of smaller mass, nor of smaller $M_r$. Convert both reactants to moles, divide each by its coefficient, and the smallest result runs out first. For 0.10 mol Mg with 0.15 mol HCl in $\text{Mg} + 2\text{HCl}$: $0.10/1 = 0.10$ against $0.15/2 = 0.075$, so HCl limits. Keep using its ratio - $n(\text{H}_2) = 0.075$ mol, not 0.15.

Both are per-mole quantities, which is why they get swapped. The Avogadro constant ($6.02 \times 10^{23}$) is a number of particles per mole and applies to any substance. The molar gas volume (24 dm^3 at r.t.p.) is a volume per mole and applies to gases only. Use $n \times 6.02 \times 10^{23}$ for a particle count, and $n \times 24$ for a gas volume.

Convert the volume before you divide: 20 g of solute in 500 cm^3 gives $20/0.5 = 40$ g/dm^3, not $20/500 = 0.04$. Then give the unit the topic actually asks for - here concentration means mass of solute per cubic decimetre, g/dm^3. A concentration in mol/dm^3 is Extended and is met later with titrations, so quoting it here answers a different question.