Question

Zinc metal reacts with hydrochloric acid to produce zinc chloride and hydrogen gas according to the equation below. What volume of hydrogen gas would be produced at 0.998 atm and 27 degrees C if 15.11 g of zinc reacted with excess hydrochloric acid?

Zn(s) + 2HCl(aq) ---> ZnCl2(aq) + H2(g)

Answer #1

Number of moles of Zn = 15.11 g / 65.38 g/mol = 0.231 mole

from the balanced equation we can say that

1 mole of Zn producs 1 mole of H2 so

0.231 mole of Zn will produce 0.231 mole of H2

PV= nRT

where, P = pressure = 0.998 atm

V = volume = ?

n = numebr of moles of H2 gas = 0.231 mole

R = Gas constant

T = temperature = 27 + 273 = 300 K

0.998 * V = 0.231 * 0.0821 *300

0.998 * V = 5.69

V = 5.69 / 0.998 = 5.70 L

Therefoe, the volume of hydrogen gas produced would be 5.70 L

A)
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equation:
Zn(s) + 2HCl(aq) ------------>
ZnCl2(aq) + H2(g)
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757 mm Hg. If the wet
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please show me the steps>>>>

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