Answer:
How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide? ... be used to produce 1.99 grams of water. 1.99 mg H2O X. 1mol H2O. 18.0g X ... c. If the reaction produces 5.3 mg of carbon dioxide how many grams of water ... X. 25mol O2. 2mol C8H18. X. 32.0g O2. 1mol O2. = 4.80 x 103g O2. Answer ...
Explanation:
Answer:
9.1 mol
Explanation:
The balanced chemical equation of the reaction is:
CO (g) + 2H2 (g) → CH3OH (l)
According to the above balanced equation, 2 moles of hydrogen gas (H2) are needed to produce 1 mole of methanol (CH3OH).
To convert 36.7 g of hydrogen gas to moles, we use the formula;
mole = mass/molar mass
Molar mass of H2 = 2.02g/mol
mole = 36.7/2.02
mole = 18.17mol
This means that if;
2 moles of H2 reacts to produce 1 mole of CH3OH
18.17mol of H2 will react to produce;
18.17 × 1 / 2
= 18.17/2
= 9.085
Approximately to 1 d.p = 9.1 mol of methanol (CH3OH).
The answer would be D because from my research it's the only one that didn't have a catalyst
Option C coz it should be ( CNH4)2. Hope i cleared your doubt
The balanced chemical reaction would be as follows:
<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
