Write the chemical equation for the following reactions and mention the type: A
ID: 933236 • Letter: W
Question
Write the chemical equation for the following reactions and mention the type: A sample of 0.556 g of an compound containing the bromide ion (Br^-) is dissolved in water and with excess of AgNO3. If the mass of the AgBr precipitate that forms g. What is the percentage by mass of Br in the original compound?? Define the following terms: Boy les law Displacemeat reaction What is the mass of magnesium chloride resulted by the reaction between hydrochloric acid and magnesium, if the volume was 540 ml, pressure 520 mmHg, and temperature 455 "F"? A sample of carbon monoxide gas occupies in 3.25 Litres at 130 Calculate the resulted temperature at which the gas will occupy 800.06 ml the pressure remains unchanged? At STP conditions, 788.89 ml of a gas weighs 0.567 g. Calculate the mass of the gas?Explanation / Answer
Q.11: When Mg reacts with HCl, H2 gas is liberated.
Given the volume, V = 540 mL = 540 mL x (1L / 1000 mL) = 0.540 L
Pressure, P = 520 mm Hg = 520 mm Hg x ( 1atm / 760 mm Hg) = 0.6842 atm
temperature, T = 455 F = 508.15 K
Now the moles of H2 gas formed can be caluclated from ideal gas equation.
PV = nRT
=> n = PV/RT = (0.6842 atm x 0.540 L) / (0.0821 L.atm.mol-1K-1 x 508.15 K) = 0.008856 mol H2
The balanced equation for the reaction of Mg and HCl is
Mg(s) + 2 HCl(aq) ---- > MgCl2(aq) + H2(g)
1 mol, -- 2 mol------------- 1 mol, ------- 1 mol
In the above balanced reation, 1 mol of MgCl2 is formed when 1 mol of H2 is liberated.
Hence moles of MgCl2 formed when 0.008856 mol H2 is liberated
= (1 mol MgCl2 / 1 mol H2) x 0.008856 mol H2 = 0.008856 mol MgCl2
Molecualas mass of MgCl2 = 95.21 g/mol
Hence mass of MgCl2 formed = 0.008856 mol x 95.21 g/mol = 0.843 g (answer)
Q:13: For STP condition, T = 273.15 K
P = 0.98692 atm
Given, volume, V = 788.89 mL = 0.78889 L
weight of the gas, W = 0.567 g
Let the molar mass be 'M'
Now applying ideal gas equation
PV = nRT = WRT / M
=> M = WRT / PV = ( 0.567g x 0.0821 L.atm.mol-1K-1 x 273.15 K) / (0.98692 atm x 0.78889 L)
= 16.22 g (answer)
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