magnesium reacts with sulphuric acid according to the equation:

Mg3 + H2SO4=MgSO4 + H2
a 250cubic centimetre aqueous solution containing 9.8g of sulphuric acid is added to 6g magnesium
a. show that the magnesium is present in excess
Does it have something to do with the unbalanced equation
b. calculate the mass of magnesium sulphate formed
c. calculate the mass of hydrogen gas measured at standard room temperature that was formed in the reaction

Where does the 3 from Mg3 come from. The equation is

Mg + H2SO4 ==> MgSO4 + H2

a. mols Mg = grams/atomic mass = ? about 0.2
mols H2SO4 = g/molar mass = ? about 0.1
Since 1 mols Mg reacts with 1 mol H2SO4 then 0.1 mols H2SO4 will use up 0,1 mol Mg and have some left over.

b. 0.1 mol H2SO4 will form 0.1 mol MgSO4. g MgSO4 = mols MgSO4 x mol mass MgSO4.

c. done the same way as part b.
Post your work if you get stuck.

a. To determine if magnesium is present in excess, we need to compare the stoichiometric ratio of magnesium to sulphuric acid in the given equation.

The balanced equation is:
3Mg + H2SO4 → MgSO4 + H2

According to the reaction equation, the stoichiometric ratio between magnesium and sulphuric acid is 3:1. This means that for every 3 moles of magnesium, we need only 1 mole of sulphuric acid to react completely.

From the given information, we know that 9.8g of sulphuric acid is present. To find the moles of sulphuric acid, we need to use its molar mass.

The molar mass of H2SO4 = (2*1) + 32 + (4*16) = 98 g/mol

So, the moles of sulphuric acid are:
moles of H2SO4 = mass/Molar mass = 9.8/98 = 0.1 mol

Now, let's calculate the moles of magnesium present using its molar mass:
molar mass of Mg = 24 g/mol

moles of Mg = mass/Molar mass = 6/24 = 0.25 mol

Comparing the moles of magnesium to sulphuric acid, we have:
Moles of magnesium / Moles of sulphuric acid = 0.25 / 0.1 = 2.5

Since the ratio is greater than 3, it indicates that there is an excess of magnesium. Therefore, the magnesium is present in excess.

b. To calculate the mass of magnesium sulphate formed, we need to use stoichiometry.

From the balanced equation:
3 moles of Mg react to form 1 mole of MgSO4

We already found that the moles of Mg are 0.25. So, the moles of MgSO4 formed will be:
moles of MgSO4 = moles of Mg × (1 mole of MgSO4 / 3 moles of Mg) = 0.25 × (1/3) = 0.083 mol

Now, let's find the molar mass of MgSO4:
molar mass of MgSO4 = 24 + 32 + (4*16) = 120 g/mol

mass of MgSO4 formed = moles of MgSO4 × Molar mass = 0.083 × 120 = 9.96 g

Therefore, the mass of magnesium sulphate formed is approximately 9.96 grams.

c. To calculate the mass of hydrogen gas formed, we need to use stoichiometry as well.

From the balanced equation:
3 moles of Mg react to form 1 mole of H2

We already found that the moles of Mg are 0.25. So, the moles of H2 formed will be:
moles of H2 = moles of Mg × (1 mole of H2 / 3 moles of Mg) = 0.25 × (1/3) = 0.083 mol

Now, let's calculate the molar mass of H2:
molar mass of H2 = 2 g/mol

mass of H2 formed = moles of H2 × Molar mass = 0.083 × 2 = 0.166 g

Therefore, the mass of hydrogen gas formed is approximately 0.166 grams.