when 1.00g of magnesium is reacted with excess hydrochloric acid, what volume of hydrogen gas will be produced at standard temperature and pressure? statndard temperature and pressure is defined as 0 C and 1 atm. At STP, 1 mole of gas occupies 22.4L.

0.93L

Explanation:

Hello,

To solve this question, we need to first of all write down the equation of reaction.

Equation of reaction,

2Mg + 4HCl → 2MgCl₂ + 2H₂

From the equation of reaction, it shows that 2 moles of Mg produces 2 moles of H₂.

Number of moles = mass / molar mass

Mass = number of moles × molar mass

Molar mass of Mg = 24g/mol

Mass = 2 × 24 = 48g

Therefore, 48g of Mg produces (2 × 22.4)L of H₂

48g of Mg = 44.8L of H₂

1g of Mg = xL of H₂

x = (1 × 44.8) / 48

x = 0.93L

1g of Mg will produce 0.93L of H₂

answer; /// i do believe that the correct answer is (a) boiling

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