1.How many grams of carbon monoxide are needed to react with an excess of iron (III) oxide to produce 198.5 grams of iron? Fe2O3(s) + 3CO(g) ----- 3CO2(g) + 2Fe(s) Show all work!
2.Hydrogen gas can be made by reacting methane (CH4) with high temperature steam: CH4(g) + H2O(g) ----- CO(g) + 3H2(g)
How many hydrogen molecules are produced when 256 grams of methane reacts with steam?
3.Ammonia (NH3) reacts with oxygen (O2) to produce nitrogen monoxide (NO) and water (H2O). Write and balance the chemical equation. How many liters of NO are produced when 2.0 liters of oxygen reacts with ammonia? Show all work!
4.Phosphoric acid reacts with sodium hydroxide: H3PO4(aq) + 3NaOH(aq) --- Na3PO4(aq) + H2O(l)
If 3.50 mol H3PO4 is made to react with 10.0 mol NaOH, identify the limiting reagent. Show all work!
Can anyone help thanks!
these are all pretty much alike. For #1,
Fe2O3 + 3CO --> 3CO2 + 2Fe
Each mole of Fe2O3 produces 2 moles of Fe
198.5g Fe = 3.373 moles Fe
So, since 3 moles CO produce 2 moles Fe, we need 3/2 * 3.373 = 5.06 moles CO
That means we need 141.677 g CO
See what you can do with the others ...
1. To determine the grams of carbon monoxide needed, we need to use the balanced equation to find the molar ratio between carbon monoxide and iron.
According to the balanced equation, the ratio between CO and Fe is 3:2. This means that for every 3 moles of CO, we will produce 2 moles of Fe.
First, we need to convert the given mass of Fe to moles using its molar mass. The molar mass of Fe is 55.85 g/mol.
198.5 g Fe * (1 mol Fe / 55.85 g Fe) = 3.55 mol Fe
Using the mole ratio, we can determine the moles of CO needed.
3.55 mol Fe * (3 mol CO / 2 mol Fe) = 5.33 mol CO
Finally, we can convert the moles of CO to grams using its molar mass. The molar mass of CO is 28.01 g/mol.
5.33 mol CO * (28.01 g CO / 1 mol CO) = 149.45 g CO
Therefore, 149.45 grams of carbon monoxide are needed to react with an excess of iron (III) oxide to produce 198.5 grams of iron.
2. To determine the number of hydrogen molecules produced, we need to use the balanced equation and convert the given mass of methane to moles.
According to the balanced equation, the ratio between CH4 and H2 is 1:3. This means that for every 1 mole of CH4, we will produce 3 moles of H2.
First, we need to convert the given mass of CH4 to moles using its molar mass. The molar mass of CH4 is 16.04 g/mol.
256 g CH4 * (1 mol CH4 / 16.04 g CH4) = 15.95 mol CH4
Using the mole ratio, we can determine the moles of H2 produced.
15.95 mol CH4 * (3 mol H2 / 1 mol CH4) = 47.85 mol H2
Finally, we can convert the moles of H2 to molecules using Avogadro's number (6.022 × 10^23 molecules/mol).
47.85 mol H2 * (6.022 × 10^23 molecules H2 / 1 mol H2) = 2.88 × 10^25 molecules of H2
Therefore, when 256 grams of methane reacts with steam, approximately 2.88 × 10^25 molecules of hydrogen are produced.
3. The balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to produce nitrogen monoxide (NO) and water (H2O) is as follows:
4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)
From the balanced equation, we can see that for every 4 moles of NH3, we produce 4 moles of NO.
To determine the number of liters of NO produced, we need to convert the given volume of oxygen to moles using the ideal gas law.
The ideal gas law is given by PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
Given that the volume of oxygen is 2.0 liters, and assuming standard conditions of temperature and pressure (STP), we can use the ideal gas law to find the number of moles of oxygen. At STP, the temperature is 273.15 K and the pressure is 1 atm.
PV = nRT
(1 atm) (2.0 L) = n (0.0821 L·atm/mol·K) (273.15 K)
n = (1 atm) (2.0 L) / (0.0821 L·atm/mol·K) (273.15 K)
= 0.0887 mol
Using the mole ratio from the balanced equation, we can determine the moles of NO produced.
0.0887 mol O2 * (4 mol NO / 5 mol O2) = 0.0709 mol NO
Finally, we can use the ideal gas law to convert the moles of NO to the volume in liters. Assuming STP, we can use the same values for R, T, and P.
PV = nRT
V = nRT / P
V = (0.0709 mol) (0.0821 L·atm/mol·K) (273.15 K) / (1 atm)
V ≈ 1.38 L
Therefore, approximately 1.38 liters of NO are produced when 2.0 liters of oxygen react with ammonia.
4. To identify the limiting reagent, we need to compare the mole ratios of the reactants from the balanced equation to the actual amounts of reactants given.
According to the balanced equation, the ratio between H3PO4 and NaOH is 1:3. This means that for every 1 mole of H3PO4, we will need 3 moles of NaOH.
Given that we have 3.50 mol of H3PO4 and 10.0 mol of NaOH, we can calculate the mole ratios of the two reactants.
Mole ratio of H3PO4 : NaOH = 3.50 mol : 10.0 mol
Mole ratio simplified = 0.35 : 1
Comparing the mole ratios, we can see that the actual mole ratio is less than the balanced equation mole ratio. This means that NaOH is present in a higher proportion than required, making it the excess reactant.
Therefore, the limiting reagent is H3PO4.
1. To determine how many grams of carbon monoxide (CO) are needed to react with an excess of iron (III) oxide (Fe2O3) to produce 198.5 grams of iron (Fe), we need to use stoichiometry.
First, let's calculate the molar mass of Fe2O3 and Fe.
Molar mass of Fe2O3: 2 * atomic mass of Fe + 3 * atomic mass of O = 2 * 55.85 g/mol + 3 * 16.00 g/mol = 159.70 g/mol
Molar mass of Fe: 55.85 g/mol
Next, we need to determine the mole ratio between Fe2O3 and CO from the balanced equation.
From the balanced equation: 1 mole of Fe2O3 reacts with 3 moles of CO.
Now, we can set up a proportion to find the moles of CO required:
(198.5 g Fe) * (1 mol Fe / 55.85 g Fe) * (3 mol CO / 1 mol Fe2O3) = x mol CO
Finally, convert the moles of CO to grams:
x mol CO * (28.01 g CO / 1 mol CO) = grams of CO needed
2. To determine how many hydrogen molecules (H2) are produced when 256 grams of methane (CH4) reacts with steam (H2O), we will again use stoichiometry.
First, calculate the molar mass of CH4 and H2.
Molar mass of CH4: 12.01 g/mol + 4 * 1.008 g/mol = 16.04 g/mol
Molar mass of H2: 2 * 1.008 g/mol = 2.016 g/mol
Next, determine the mole ratio between CH4 and H2 from the balanced equation.
From the balanced equation: 1 mole of CH4 reacts to produce 3 moles of H2.
Now, set up a proportion to find the moles of H2 produced:
(256 g CH4) * (1 mol CH4 / 16.04 g CH4) * (3 mol H2 / 1 mol CH4) = x mol H2
Finally, convert the moles of H2 to molecules:
x mol H2 * (6.022 x 10^23 molecules H2 / 1 mol H2) = number of H2 molecules produced
3. The balanced chemical equation for the reaction between ammonia (NH3) and oxygen (O2) to produce nitrogen monoxide (NO) and water (H2O) is:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
To determine how many liters of NO are produced when 2.0 liters of oxygen reacts with ammonia, we need to use the mole ratio between O2 and NO from the balanced equation.
From the balanced equation: 5 moles of O2 react to produce 4 moles of NO.
First, convert 2.0 liters of oxygen to moles using the ideal gas law or the fact that 1 mole of any ideal gas occupies 22.4 liters at standard temperature and pressure (STP):
2.0 L O2 * (1 mol O2 / 22.4 L O2) = x mol O2
Now, use the mole ratio to find the moles of NO produced:
x mol O2 * (4 mol NO / 5 mol O2) = x mol NO
Finally, convert the moles of NO to liters using STP:
x mol NO * (22.4 L NO / 1 mol NO) = liters of NO produced
4. To identify the limiting reagent when 3.50 mol of H3PO4 reacts with 10.0 mol of NaOH, we need to compare the stoichiometric ratios of the reactants.
From the balanced equation: 1 mole of H3PO4 reacts with 3 moles of NaOH.
First, determine the ratio of moles of H3PO4 to moles of NaOH:
3.50 mol H3PO4 / 10.0 mol NaOH = x mol H3PO4 / 3 mol NaOH
If the calculated ratio is greater than 1, then NaOH is the limiting reagent. If the ratio is less than 1, then H3PO4 is the limiting reagent.