Question

A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams of methanol are produced when 2. 8 grams of carbon monoxide reacts with 0. 50 grams of hydrogen gas

Answer 1

To find the grams of methanol produced in a synthesis reaction, we must use the stoichiometry of the reaction. Convert the given amounts of carbon monoxide (CO) and hydrogen gas (H2) into moles, identify the limiting reactant, calculate the moles of CH3OH produced using the stoichiometric ratio, and then convert the moles of CH3OH into grams using its molar mass.

Step-by-step explanation:

In the synthesis reaction:

CO + H2 → CH3OH

To find the grams of methanol produced, we need to use the stoichiometry of the reaction. First, we convert the given amounts of carbon monoxide (CO) and hydrogen gas (H2) into moles using their molar masses.

2.8 g CO * (1 mol CO / molar mass of CO) = x mol CO

0.50 g H2 * (1 mol H2 / molar mass of H2) = y mol H2

Next, we identify the limiting reactant by comparing the moles of CO and H2. The reactant with fewer moles is the limiting reactant. Assume CO is limiting and use the mole ratio from the balanced equation to calculate the moles of CH3OH produced.

x mol CO * (1 mol CH3OH / 1 mol CO) = z mol CH3OH

Finally, convert the moles of CH3OH into grams using its molar mass.

z mol CH3OH * (molar mass of CH3OH / 1 mol CH3OH) = grams of CH3OH

Substitute the values into the equation to find the grams of methanol produced:

Answer 2

The synthesis reaction between CO and H2 to form CH3OH is used to calculate that 3.204 grams of methanol are produced from 2.8 grams of carbon monoxide and 0.50 grams of hydrogen, using stoichiometry based on the balanced chemical equation.

Step-by-step explanation:

The student has asked about a synthesis reaction where carbon monoxide (CO) reacts with hydrogen gas (H2) to produce methanol (CH3OH). To calculate the amount of methanol produced from 2.8 grams of carbon monoxide and 0.50 grams of hydrogen, we need to use stoichiometry based on the balanced chemical equation:

`CO(g) + 2H2(g) \rightarrow CH_3OH(g)`

First, we find the molar mass of CO and H2, which are 28.01 g/mol and 2.02 g/mol, respectively. Then we convert the mass of each reactant to moles:

• 2.8 g CO * (1 mol CO / 28.01 g CO) = 0.1 mol CO
• 0.50 g H2 * (1 mol H2 / 2.02 g H2) = 0.2475 mol H2

According to the balanced equation, 1 mole of CO reacts with 2 moles of H2 to produce 1 mole of CH3OH. The limiting reactant is CO, as it would be completely consumed before H2. Therefore, the maximum amount of CH3OH produced would be based on the amount of CO.

To find the mass of CH3OH produced:

0.1 mol CO * (1 mol CH3OH / 1 mol CO) * (32.04 g CH3OH / 1 mol CH3OH) = 3.204 g CH3OH

Hence, 3.204 grams of methanol are produced when 2.8 grams of carbon monoxide reacts with 0.50 grams of hydrogen gas.