If a 45.0-mL sample of gas at 26.5 degree C is heated to 55.2 degree C, what is
ID: 951585 • Letter: I
Question
If a 45.0-mL sample of gas at 26.5 degree C is heated to 55.2 degree C, what is the new volume of the gas sample (at constant pressure)? For each of the following sets of volume/temperature limiting reactant If water is added to magnesium nitride, ammonia gas is produced when the mixture is heated. Mg_3 N_2 (s) + 3H_2 O(l) rightarrow 3 MgO (s) + 2NH_3 (g) If 10.3 g of magnesium nitride is treated with water, what volume of ammonia gas would be collected at 24 degree C and 752 mm Hg? What volume does a mixture of 14.2 g of He and 21.6 g of H_2 occupy at 28 degree C and 0.985 atm? A sample of hydrogen gas has a volume of 145 mL when measured at 44 degree C and 1.47 atm. What volume When calcium carbonate is heated strongly, carbon dioxide gas is released CaCO_3 (s) rightarrow CaO(s) + CO_2 (g) What volume of CO_2 (g), measured at STP, is produced if 15.2 g of CaCO_3 (s) is heated?Explanation / Answer
13. Using, V1/T1 = V2/T2
with,
V1 = 45 ml
T1 = 26.5 oC
V2 = ?
T2 = 55.2 oC
we get,
volume occupied by gas at 55.2 oC, V2 = 45 x 55.2/26.5 = 93.73 ml
46. CaCO3(s) ---> CaO(s) + CO2(g)
1 mole of CaCO3 gives 1 mole of CO2
moles of CaCO3 = 15.2 g/100.087 g/mol = 0.152 mols
moles of CO2 produced = 0.152 mols
Volume of CO2 produced = 0.152 x 22.4 = 3.40 L
49. Mg3N2(s) ---> 3MgO(s) + 2NH3(g)
moles of Mg3N2 = 10.3 g/100.9494 g/mol = 0.102 mols
moles of NH3 produced = 2 x 0.102 = 0.204 mols
Volume of NH3 produced = nRT/P = 0.204 x 0.08206 x (24 + 273)/(752/760) = 5.025 L
50. moles of He = 14.2 g/4 = 3.55 mols
moles of H2 = 21.6 g/2 = 10.8 mols
Total moles (He+H2) = 14.35 mols
Volume occupied by gas = nRT/P = 14.35 x 0.08205 x (28 + 273)/0.985 = 359.80 L