2NO(g) + 5H2(g) ⟶ 2NH3(g) + 2H2O(g)If 45.4 grams of NO and 12.1 grams of H2 react, what is the largest amount of ammonia that can be formed? The final answer should be reported to one place after the decimal point

25.8g of NH3 can be formed

Explanation:

To solve this question we must convert the mass of each reactant to moles using molar mass. With the chemical equation we can find limiting reactant. The moles of limiting reactant are used yo find the moles of ammonia produced and its mass:

Moles NO -30.01g/mol-

45.4g * (1mol / 30.01g) = 1.51 moles

Moles H2 -Molar mass: 2.01g/mol-

12.1g * (1mol / 2.01g) = 6.02 moles

For a complete reaction of 6.02 moles of H2 are needed:

6.02mol H2 * (2mol NO / 5mol H2) = 2.41 moles NO are needed

As there are just 1.51 moles, NO is limiting reactant

Moles NH3:

1.51 moles NO * (2mol NH3 / 2mol NO) = 1.51 moles NH3

Mass NH3 -Molar mass: 17.031g/mol-

1.51 moles NH3 * (17.031g / mol) =