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When the balanced reaction equation is:
P4O10 + 6H2O→ 4 H3PO4
when we have the mass of P4O10 = 10 g and the molar mass of P4O10=284 g/mol & we have the molar mass of H3PO4 =98 g/mol so we can get the mass of H3PO4 by substitution by:
mass of H3PO4 = (mass of P4O10)/(molar mass of P4O10) * 4(mol of H3PO4)*molar mass of H3PO4
∴mass of H3PO4 = (10 / 284) * 4 * 98 = 13.8 g
Answer:
222.91g of KCl
Explanation:
First, we need to write a balanced equation for the reaction. This is illustrated below
2K + Cl2 —> 2KCl
Molar Mass of KCl = 74.55 g/mol
Mass of KCl from the balanced equation = 2 x 74.55 = 149.1g
Molar Mass of Cl2 = 2 x 35.45 = 70.9g.
From the equation,
70.9g Cl2 produced 149.1g of KCl.
Therefore, 106 g of Cl2 will produce = (106 x 149.1)/70.9 = 222.91g of KCl
The mass of 45.0 L of Cl₂ at 87.0° C and 950 mm Hg is 134.7214 g.
Volume = 45.0 L
Temperature = = (87.0 + 273) K = 360 K
Pressure = 950 mm Hg (1 mm Hg = 0.00131579 atm) = 1.25 atm
The formula used to calculate moles is as follows.
∴PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into the above formula as follows:
∴ PV = nRT
=> 1.25 atm × 45.0 L = n × 0.0821 L atm/mol K × 360K
=> n = 1.25 atm × 45.0 L / n × 0.0821 L atm/mol K × 360K
=> n = 56.25 / 29.556 mol
=> n = 1.90 mol
Moles is the mass of a substance divided by its molar mass. So, the mass of Cl₂ (molar mass = 70.906 g/mol) is calculated as follows:
∴ Moles = mass / molar mass
=> 1.90 mol = mass / 70.906 g/mol
=> mass = 134.7214 g
Thus, we can conclude that the mass of 45.0 L of Cl₂ at 87.0° C and 950 mm Hg is 134.7214 g.
To know more about the Ideal Gas Law :
brainly.com/question/28976906
