For the reaction 2 nh3 + ch3oh → products, how much ch3oh is needed to react with 93.5 g of nh3? 1. 1.31 mol 2. 46.8 mol 3. 2.75 mol 4. 11.3 mol 5. 3.32 mol 6. 88.1 mol 7. 5.50 mol

3) 2.75 mol

Explanation:

2 nh3 + ch3oh → products

From the equation above, it can be deduced that 2 moles of NH3 reacts with 1 mole of CH3OH to form the products.

Since the answer is needed in moles,

First we convert g of NH3 to moles. The molar mass of NH3, using the atomic weights on the periodic table, = N (14.0) + 3H (3 x 1.0) = 17.0 g/mole.

93.5 g NH3 x (1 mole NH3 / 17.0 g NH3) = 5.5 moles NH3

The equation tells us that 2 moles of NH3 reacts with 1 mole of CH3OH.

2 moles of NH3 = 1 mole of CH3OH

5.5 moles of NH3 = x mole of CH3OH

x mole = (5.5 moles of NH3 * 1 mole of CH3OH) / 2 moles of NH3

x = 2.75 mol

you need to give more information

indium, and thallium have three electrons i think