How many grams of magnesium oxide can be “dissolved” in 4.9 grams of sulfuric acid?

We will carry out the solution, draw up an equation, select the coefficients:
MgO + H2SO4 = MgSO4 + H2O – ion exchange reaction, magnesium sulfate and water are released;
Let’s calculate the molar masses and the substances that react:
M (MgO) = 24.3 + 16 = 40.3 g / mol;
M (H2SO4) = 1 * 2 + 32 + 16 * 4 = 98 g / mol.
Determine the amount of moles of sulfuric acid:
Y (H2SO4) = m / M = 4.9 / 98 = 0.05 mol.
Let’s make the proportion:
0.05 mol (H2SO4) – X mol (MgO);
-1 mol – 1 mol from here, X mol (MgO) = 0.05 * 1/1 = 0.05 mol.
Let’s calculate the mass of magnesium oxide:
m (MgO) = Y * M = 0.05 * 40.3 = 2.015 g.
Answer: the mass of the oxide is 2.015 g.