Question

Consider the reaction 2 nh3 + ch3oh → products. how much nh3 is needed to react exactly with 21 grams of ch3oh?

Answer 1

Approximately 22.32 grams of NH3 are required to react with 21 grams of CH3OH, using the molar mass of NH3 and CH3OH and the stoichiometric ratios from the balanced chemical equation.

Step-by-step explanation:

To determine how much NH3 is needed to react exactly with 21 grams of CH3OH, we need to use stoichiometry. First, we need the molecular weights of NH3 and CH3OH to convert grams to moles. The molecular weight of NH3 (ammonia) is about 17.03 g/mol, and the molecular weight of CH3OH (methanol) is approximately 32.04 g/mol.

Using the provided molecular weights, you can calculate the moles of CH3OH by dividing the mass by the molecular weight:

moles of CH3OH = 21 g / 32.04 g/mol ≈ 0.655 moles

The balanced chemical equation 2 NH3 + CH3OH → products tells us that 1 mole of CH3OH reacts with 2 moles of NH3. Hence, for 0.655 moles of CH3OH, we need:

0.655 moles of CH3OH x 2 moles of NH3 / 1 mole of CH3OH = 1.31 moles of NH3

To find the mass of NH3 required, multiply the moles of NH3 needed by its molecular weight:

mass of NH3 = 1.31 moles x 17.03 g/mol ≈ 22.32 grams

Therefore, approximately 22.32 grams of NH3 are required to react with 21 grams of CH3OH.

Answer 2

The NH3 needed to react exactly with 21 grams of CH3OH is calculated as follows
2NH3 +CH3OH = products
find the moles of CH3OH used

moles = mass/molar mass
= 21 grams/32 g/mol = 0.656 moles


from the mole ratio between NH3 to CH3OH which is 2:1 the moles of NH3 is therefore = 0.656 x2/1 = 1.312 moles

mass of NH3 is therefore= moles of NH3 x molar mass of NH3

= 1.312 moles x 17 g/mol= 22.3 grams