CO + 2 H2 → CH3OH
<span> find # of mols in each reactants, </span>
<span>152500 g CO x 1 mol CO / 28.01g CO = 5444 mol CO </span>
<span>24500 g H2 x 1 mol H2 / 2.02 g H2 = 12129 mol H2 </span>
<span>mol ratio between CO and H2 is 1:2, which means each mol of production of CH3OH need 1 mol of CO and 2 mol of H. </span>
<span>H2 is enough to produce 6064 mols of CH3OH but there are only 5444mol of CO. </span>
<span>5444 mol CH3OH x molar mass of CH3OH / 1 mol CH3OH </span>
<span>= 174371 g = 174.4 kg</span>
Answer:
d. oxygen (O2) is the answer
Answer:
Explanation:
Hello there!
In this case, according to the given chemical reaction, it is possible for us to relate the rate of formation of C and the rate of consumption of B as shown below:
Thus, we solve for the rate of change of B as shown below:
Best regards!
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
Answer:
The answer will result in one significant digit.
Explanation:
1. Subtract:
12.1 - 12.004 = 0.096
2. ROund to the correct number of significant digits:
For addition and subtraction we round the sum or the difference to the least number of digits after the decimal point there are in the problem.
For example, SInce 12.1 has one number after the decimal point and 12.004 has three numbers after the decimal point, we round the answer to one number after the decimal point because 12.1 is the number in the problem which has the least amount of digits after the decimal point.
We round 0.096 to 0.1.
0.1 has one significant digit.
