Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
Answer:Macaya (floruit 1802), was a Kongolese-born Haitian revolutionary military leader. Macaya was one of the first black rebel leaders in Saint-Domingue to ally himself with the French Republican commissioners Sonthonax and Polverel. He helped to lead forces that recaptured Cap-Français on behalf of the French Republicans.
Macaya was born in west-central Africa, probably in the Kingdom of Kongo, and taken to the French colony of Saint-Domingue as a slave. After the outbreak of the 1791 slave rebellion in northern Saint-Domingue, Macaya became a lieutenant of an elderly rebel commander named Pierrot. Pierrot's rebel forces were based in the hills outside of Le Cap (Cap Francaise), near Bréda plantation by 1793.
Explanation:
3.52g BiCl3 × 1 mol BiCl3/ 315.34g BiCl3 × 3 mol Cl/ 2 mol BiCl3 × 70.906g Cl/ 1 mol Cl= 1.187 g Cl
The balanced chemical reaction is
<span>2al + 3cl2 = 2alcl3
To determine the maximum amount of product, we need to determine which is the limiting reactant. Then, use the initial amount of that reactant to calculate the amount of the product that would be produced. We do as follows:
7 mol Al (3 mol Cl2 / 2 mol Al) = 10.5 mol Cl2
8 mol Cl2 ( 2 mol Al / 3 mol Cl2) = 5.3 mol Al
Thus, it is Cl2 that is the limiting reactant.
8 mol Cl2 ( 2 mol AlCl3 / 3 mol Cl2) = 5.3 moles of AlCl3 is produced</span>
