Answer:
22.77 g.
he limiting reactant is O₂, and the excess reactant is Mg.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (16.3 g) of Mg and (4.52 g) of oxygen:
no. of moles of Mg = mass/molar mass = (16.3 g)/(24.3 g/mol) = 0.6708 mol.
no. of moles of O₂ = mass/molar mass = (4.52 g)/(16.0 g/mol) = 0.2825 mol.
So. 0.565 mol of Mg reacts completely with (0.2825 mol) of O₂.
<em>∴ The limiting reactant is O₂, and the excess reactant is Mg (0.6708 - 0.565 = 0.1058 mol).</em>
<u><em>Using cross multiplication:</em></u>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.565 mol of Mg produce → <em>0.565 mol of MgO.</em>
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.565 mol)(40.3 g/mol) = <em>22.77 g.</em>
Answer:
atoms are neutral or they have equal number of proton and electron but ions can be cation(positively charged) or anion(negatively charged) based on the number of electron the atm lose and gain respectively
Answer : The mass of helium gas added must be 12.48 grams.
Explanation : Given,
Mass of helium (He) gas = 6.24 g
Molar mass of helium = 4 g/mole
First we have to calculate the moles of helium gas.
Now we have to calculate the moles of helium gas at doubled volume.
According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,
or,
where,
= initial volume of gas = V
= final volume of gas = 2V
= initial moles of gas = 1.56 mole
= final moles of gas = ?
Now we put all the given values in this formula, we get
Now we have to calculate the mass of helium gas at doubled volume.
Therefore, the mass of helium gas added must be 12.48 grams.
Answer:
10/3 m/s. v=d/t. so 100/30 = 10/3
