A2.00−g sample of a metal x (that is known to form x2+ ions) was added to 0.100 l of 0.500 m h2so4. after all the metal had reacted, the remaining acid required 0.0543 l of 0.500 m naoh solution for neutralization. calculate the molar mass of the metal and identify the element.

87.5 g/mol, Sr (strontium)

Explanation:

The reaction between the metal and the acid is:

X + H₂SO₄ → 2H⁺ + XSO₄ (ion sulfate is SO₄⁻²)

The initial number of moles of the acid is the concentration multiplied by the volume:

n = 0.500*0.1 = 0.05 mol

The reaction between the acid and the base is:

H₂SO₄ + NaOH → NaHSO₄ + H₂O

The number of moles of NaOH is:

0.0543x0.5 = 0.02715 mol

Which is the same number of moles of H₂SO₄, because of the stoichiometry 1:1. So, in the first reaction, reacted:

0.05 - 0.02715 = 0.02285 mol of H₂SO₄

And, for the stoichiometry, there was 0.02285 mol of X. The molar mass of X is the mass divided by the number of moles:

2/0.02285 = 87.5 g/mol

For the periodic table, X is Sr,  strontium.

pie chart because it is in percents ( % )