How many grams of magnesium oxide would be produced if 3.8 grams of magnesium reacted completely with 2.1 grams of oxygen

5.3g of MgO would be produced

Explanation:

Based on the chemical reaction:

2Mg(s) + O₂(g) → 2MgO(s)

1 mole of Mg reacts per mole of oxygen.

To solve this question we need to convert the mass of each reactant to moles in order to find limiting reactant using the chemical equation:

Moles Mg -Molar mass: 24.305g/mol-:

3.8g Mg * (1mol / 24.305g) = 0.1563moles Mg

Moles O₂ -Molar mass: 32g/mol-:

2.1g O₂ * (1mol / 32g) = 0.0656 moles O₂

For a complete reaction of 0.0656 moles of O₂ are required:

0.0656 moles O₂ * (2 mol Mg / 1mol O₂) = 0.132 moles Mg

As there are 0.1563 moles of Mg, Mg is the excess reactant and O₂ the limiting reactant.

0.0656 moles O₂ produce:

0.0656 moles O₂ * (2 mol MgO / 1mol O₂) = 0.132 moles MgO

The mass is -Molar mass MgO: 40.3g/mol-:

0.132 moles MgO * (40.3g / mol) =

5.3g of MgO would be produced

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles