If 4.0 g of H₂ are made to react with excess CO, 31.7 g CH₃OH should be formed. If 28.0 g of CH₃OH are actually produced, what is the percent yield? CO(g) + 2 H2(g) CH3OH(l) *

87.5 % is the percent yield

Explanation:

We convert the mass of H₂ to moles: 4 g / 2g/mol = 2 moles of H₂

The reaction is: CO(g) + 2H₂(g) → CH₃OH(l)

So ratio is 2:1. Therefore, If I have 2 moles of hydrogen I would produce 1 mol of methanol.

So, as molar mass of methanol is 32g/mol; 32 g is the mass of produced methanol, at the 100 % yield reaction (theoretical yield)

Percent yield will be: (produced yield/ theoretical yield) . 100

(28 g / 32g) . 100 = 87.5 %

The percent yield of the reaction is 88.3 %

Explanation:

Step 1: Data given

Mass of H2 = 4.0 grams

Mass of CH3OH produced = 31.7 grams

Mass of CH3OH actually produced = 28.0 grams

Step 2: The balanced equation

CO(g) + 2 H2(g) → CH3OH(l)

Step 3: Calculate the percent yield

Percent yield = (actual mass CH3OH / theoretical mass CH3OH) * 100 %

Percent yield = (28.0 grams / 31.7 grams) * 100%

Percent yield = 88.3 %

The percent yield of the reaction is 88.3 %

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.