?RSI LABORATORY ASSIGNMENT. 1. A 0.215 g sample of zine metal reacted with exces
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Question
?RSI LABORATORY ASSIGNMENT. 1. A 0.215 g sample of zine metal reacted with excess hydrochloric acid accordin ing balanced chemical reaction: g to the follow- Zn(s)+2 HCa)ZnCl(ag)H(8) The volume of hydrogen gas collected over water was 80.5 mL. at 20 °C and a barometric pres- sure of 756 mmHg. Calculate the molar volume at STP for hydrogen. ?.al T. 20e-A3n P-lar -0.02 23.5% 1e254 as -24.4L A 0.136 g sample of an unknown metal (X) reacted with excess hydrochloric acid according to the following balanced chemical reaction: 2. 6 HCI(aq) ? 2XCIy(aq) + 3H2(g) 2X(s) + The volume of hydrogen gas collected over water was 90.5 mL at 20 °C and a barometric pres- sure of 756 mmHg. Determine the molar mass of the unknown metal. Use the Periodic Table ofelements to identify the metal· V-qo.SmL-20.1 -2.50& asortio, sszzeglnopereran18.genGaslonhnoon.fe PEARSONExplanation / Answer
1) NO of mol of Zn reacted = 0.215/65.37 = 0.0033 mol
from equation, 1 mol Zn = 1 mol H2
NO of mol of H2 liberated = 0.0033 mol
at 20c, vapor pressure of water = 17.5 torr
partial pressure of H2 liberated(P1) = 756- 17.5 = 738.5 torr
= 738.5/760
= 0.972 atm
T1 = 20c = 293 k
V1 = 80.5 ml = 0.0805 L
at STP, P2 = 1 atm
T2 = 273 k
v2 = ?
P1V1/T1 = P2V2/T2
(0.972*0.0805/293) = (1*V2/273)
v2 = 0.073 L
molar volume of H2 = V/n
= 0.073/0.0033
= 22.12 mol/l
2)
partial pressure of H2 liberated(P) = 756- 17.5 = 738.5 torr
= 738.5/760
= 0.972 atm
T = 20c = 293 k
V = 90.5 ml = 0.0905 L
no of mol of H2 = PV/RT
= (0.972*0.0905)/(0.0821*293)
= 0.00365 MOL
FROM equation, 2 mol X = 3 mol H2
No of mol of X reacted = 0.00365*2/3 = 0.00243 mol
Molarmass of X = W/n
= 0.136/0.00243
= 56 g/mol
unknown metal = x = Fe
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