Answer:
The correct answer is: 6.6 g MgO
Explanation:
First we have to write and balance the chemical reaction as follows:
2Mg(s) + O₂(g) → 2MgO(s)
That means that 2 moles of Mg(s) react with 1 mol of O₂(g) to give 2 moles of MgO(s). If Mg is totally consumed and a mass of O₂ remains unaltered after reaction, t<em>he limiting reactant is Mg</em>. We use the limiting reactant to calculate the mass of product.
According to the balanced chemical equation, 2 moles of Mg(s) produce 2 moles of MgO(s).
2 moles Mg = 2 mol x molar mas Mg= 2 mol x 24.3 g/mol = 48.6 g Mg
2 moles MgO= 2 mol x (molar mass Mg + molar mass O) = 2 mol x (24.3 g/mol + 16 g/mol) = 80.6 g MgO
The stoichiometric ratio is 80.6 g MgO/48.6 g Mg. So, we multiply this ratio by the mass of consumed Mg (4.0 g) in order to obtain the produced mass of MgO:
4.0 g Mg x 80.6 g MgO/48.6 g Mg = 6.63 g MgO
6.6 grams of magnesium oxide are formed.
Answer:
Using the formula cards again, add the coefficient of 2 in front of the formula and have them recalculate the number of each element and the total number of atoms in each element.
Explanation:
Answer:
D
Explanation:
Heat sometimes gets "lost" when work is done
Answer:
333.7g of antifreeze
Explanation:
Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:
ΔT = Kf × m × i
Where:
ΔT is change in temperature (0°C - -20°C = 20°C)
Kf is freezing point depression constant (1.86°C / m)
m is molality of solution (moles solute / 0.5 kg solvent -500g water-)
i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)
Replacing:
20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1
5.376 = moles solute
As molar mass of ethylene glycol is 62.07g/mol:
5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.
