a quantity of .225g of a metal M (molar mass = 27.0 g/mol) liberated .303 L of m

Question

a quantity of .225g of a metal M (molar mass = 27.0 g/mol) liberated .303 L of molecular hydrogen(measured at 17 degrees C and 741 mmHg) from an excess of hydrochloric acid. Deduce from these data the corresponding equation and write the formulas for the oxide and sulfate of M. Using the smallest whole number coefficients write the balanced equation for the reaction of the metal M with hydrochloric acid. Include the phase abbreviations. What is the formula of oxide M? WHat is the formula of the sulfate of M?

Explanation / Answer

moles metal = 0.225 g/ 27.0 g/mol=0.00833

741/760=0.975 atm
T = 17 + 273 = 290 K
moles H2 = 0.303 x 0.975 / 0.08206 x 290=0.0124

the ratio between H2 and M is 0.0124/ 0.00833 = 1.5

1.5 mol of H2 for each mole of M
3.0 mol H2 for 2 moles of M

2 M + 6 HCl = 2 MCl3 + 3 H2

M2O3
M2(SO4)3

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