How do I find the amount of salt produced from 10.0 g of MgCl2 reacting with H2SO4? The balanced equation is MgCl2+H2SO4-->MgSO4+2HCl?

Amount of salt.....

MgCl2(aq) + H2SO4(aq) --> No reaction

MgSO4 is soluble in water, and so is HCl. There is no insoluble product and therefore, no reaction.

You can go through the motions of calculating the mass of a (nonexistent) product, but wouldn't it have been better if your teacher had actually paid attention to the solubility rules and picked compounds that would actually give an insoluble product. (The solubility of MgSO4 is 35.1 g/100 mL at 20 °C.) Your teacher could just as easily chosen barium chloride, BaCl2. Barium sulfate is quite insoluble in water. Impress your teacher with your knowledge of the solubility rules and turn in this problem.

BaCl2(aq) + H2SO4(aq) --> BaSO4(s) + 2HCl(aq)
10.0g ................ ................... ???g

10.0g BaCl2 x (1 mol BaCl2 / 208.2g BaCl2) x (1 mol BaSO4 / 1 mol BaCl2) x (236.36g BaSO4 / 1 mol BaSO4) = 11.4 g BaSO4

The problem is set up the same way using the unit-factor method whether its BaCl2 or MgCl2. The only differences are the molar masses of BaCl2 and BaSO4 vs MgCl2 and MgSO4.

Molar mass of MgCl₂ = (24.3 + 35.5×2) g/mol = 95.3 g/mol
Molar mass of MgSO₄ = (24.3 + 32.1 + 16.0×4) g/mol = 120.4 g/mol

MgCl₂ + H₂SO₄ → MgSO₄ + 2HCl
Mole ratio MgCl₂ : MgSO₄ = 1 : 1

No. of moles MgCl₂ reacted = (10.0 g) / (95.3 g/mol) = 0.105 mol
No. of moles of MgSO₄ producted = 0.105 mol
Mass of MgSO₄ produced = (0.105 mol) × (120.4 g/mol) = 12.6 g

(Actually, there is no reaction between MgCl₂ and H₂SO₄.)

Amount of salt?? moles or grams??
find molar mass MgCl2 = A
find molar mass MgSO4 = B

(10 / A) X B << use a calculator

sry i meant how do i find the mass of salt as in grams

Molar mass of MgCl2 = 95.211 g/mol, so you have 10.0 g / 95.211 g/mol = 0.105 moles of MgCl2.

The balanced equation:
MgCl2+H2SO4-->MgSO4+2HCl
shows that for each mole of MgCl2 reacted, you get 1 mole of MgSO4 product.

Molar mass of MgSO4 = 120.366 g/mol and you have 0.105 moles, so
0.105 mole * 120.366 g/mol = 12.6 g of MgSO4.