To solve this problem, we need to follow these steps:

1. First, we need to determine the number of moles of magnesium. We know that the molar mass of magnesium (Mg) is approximately 24.31 g/mol. So, the number of moles of magnesium (n) can be calculated using the formula: n = mass / molar mass. Substituting the given mass of magnesium (1.00 g), we get: n = 1.00 g / 24.31 g/mol = 0.0412 mol.

2. From the balanced chemical equation, we can see that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas. Therefore, the number of moles of hydrogen gas produced is equal to the number of moles of magnesium, which is 0.0412 mol.

3. At SATP (Standard Ambient Temperature and Pressure), one mole of any gas occupies a volume of 24.8 L (or 24800 cm³). Therefore, the volume of hydrogen gas produced can be calculated by multiplying the number of moles of hydrogen gas by the molar volume of gas at SATP. So, the volume of hydrogen gas (V) is: V = n * molar volume = 0.0412 mol * 24800 cm³/mol = 1021.76 cm³.

So, the volume of the balloon filled with the hydrogen gas produced at SATP is approximately 1022 cm³.

Question

To solve this problem, we need to follow these steps:

1. First, we need to determine the number of moles of magnesium. We know that the molar mass of magnesium (Mg) is approximately 24.31 g/mol. So, the number of moles of magnesium (n) can be calculated using the formula: n = mass / molar mass. Substituting the given mass of magnesium (1.00 g), we get: n = 1.00 g / 24.31 g/mol = 0.0412 mol.

2. From the balanced chemical equation, we can see that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas. Therefore, the number of moles of hydrogen gas produced is equal to the number of moles of magnesium, which is 0.0412 mol.

3. At SATP (Standard Ambient Temperature and Pressure), one mole of any gas occupies a volume of 24.8 L (or 24800 cm³). Therefore, the volume of hydrogen gas produced can be calculated by multiplying the number of moles of hydrogen gas by the molar volume of gas at SATP. So, the volume of hydrogen gas (V) is: V = n * molar volume = 0.0412 mol * 24800 cm³/mol = 1021.76 cm³.

So, the volume of the balloon filled with the hydrogen gas produced at SATP is approximately 1022 cm³.

Knowee AI · Accepted Answer

To solve this problem, we need to follow these steps:

1. First, we need to determine the number of moles of magnesium. We know that the molar mass of magnesium (Mg) is approximately 24.31 g/mol. So, the number of moles of magnesium (n) can be calculated using the formula: n = mass / molar mass. Substituting the given mass of magnesium (1.00 g), we get: n = 1.00 g / 24.31 g/mol = 0.0412 mol.

2. From the balanced chemical equation, we can see that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of hydrogen gas. Therefore, the number of moles of hydrogen gas produced is equal to the number of moles of magnesium, which is 0.0412 mol.

3. At SATP (Standard Ambient Temperature and Pressure), one mole of any gas occupies a volume of 24.8 L (or 24800 cm³). Therefore, the volume of hydrogen gas produced can be calculated by multiplying the number of moles of hydrogen gas by the molar volume of gas at SATP. So, the volume of hydrogen gas (V) is: V = n * molar volume = 0.0412 mol * 24800 cm³/mol = 1021.76 cm³.

So, the volume of the balloon filled with the hydrogen gas produced at SATP is approximately 1022 cm³.

1.00 g of magnesium ribbon reacted with 25.0 cm3 of 1.00 mol dm–3 hydrochloric acid. Determine the volume, in cm3, of a balloon filled with the hydrogen gas produced at SATP (25°C and 100 kPa). Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

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