A chemist uses hot hydrogen gas to convert chromium(III) oxide to pure chromium. How many grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, Cr203?
Cr2O3 + H2Cr + H2O

3.0242 g of H2

Explanation:

The balanced equation for the reaction is given as;

Cr2O3 + 3H2 → 2Cr + 3H2O  

From the reaction, the mole ratio between Cr2O3 and H2 is given as;

1 mol of Cr2O3 reacts with 3 mol of H2

Converting to mass using;

Mass = Molar mass * Number of moles

Cr2O3;

Mass = 151.99 g/mol * 1 mol = 151.99 g

H2;

Mass = 2.016 g/mol * 3 mol = 6.048 g

How many grams of hydrogen are needed to convert 76  grams of chromium(III) oxide, Cr203?

151.99 = 6.048

76 = x

x = 76 * 6.048 / 151.99

x = 3.0242 g

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.