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To determine the limiting reagent, you need to find the molar amounts of each reactant. Compare the molar amounts of P and I to the balanced chemical equation to see which one is present in the lower stoichiometric amount. The reactant that gives the smaller amount of product is the limiting reagent.
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If 638.44 g CuSO4 reacts with 240.0 g NaOH which is the limiting reagent?
Sodium hydroxide is the limiting reagent.
What is the limiting reagent when 150.0g of nitrogen react with 32.1 g of hydrogen?
To determine the limiting reagent, calculate the moles of each reactant: 150.0g nitrogen is 5.36 moles and 32.1g hydrogen is 31.8 moles. Using the balanced chemical equation, you can see that nitrogen is the limiting reagent because it will be completely consumed before all the hydrogen is reacted.
What is the limiting reagent when 0.343 g of magnesium chloride is reacted with 10.0 mL of 2.00 M sodium hydroxide?
To determine the limiting reagent, calculate the number of moles of each reactant: 0.343 g of MgCl2 and 10.0 mL of 2.00 M NaOH. Convert 0.343 g of MgCl2 to moles and 10.0 mL of 2.00 M NaOH to moles to see which reactant is present in the lower stoichiometric amount. Whichever reactant yields the lower amount of product is the limiting reagent.
Identify the limiting reagent and the volume of CO2 formed when 11 L CS2 reacts with 18 L O2 to produce CO2 gas and SO2 gas at STP. CS2(g) plus 3O2(g) and rarr CO2(g) plus 2SO2(g)?
To determine the limiting reagent, we need to compare the amount of CO2 that could be produced from each reactant. First, convert the volumes to moles using the ideal gas law. Then, use the coefficients from the balanced chemical equation to find the amount of CO2 that each reactant could produce. The limiting reagent is the one that produces the least amount of CO2. The volume of CO2 formed can be calculated using the stoichiometry of the limiting reagent.
If 0.25 g of Aluminum react with 0.6g of Copper sulfate then which one is the limiting reactant?
To determine the limiting reactant, we need to compare the moles of each reactant. First, calculate the moles of aluminum and copper sulfate separately. Then, determine the mole ratio between them and see which reactant is present in lower amount compared to the stoichiometric ratio. The reactant that is present in lower moles is the limiting reactant.
If 638.44 g CuSO4 reacts with 240.0 g NaOH which is the limiting reagent?
Sodium hydroxide is the limiting reagent.
What is the limiting reagent when 150.0g of nitrogen react with 32.1 g of hydrogen?
To determine the limiting reagent, calculate the moles of each reactant: 150.0g nitrogen is 5.36 moles and 32.1g hydrogen is 31.8 moles. Using the balanced chemical equation, you can see that nitrogen is the limiting reagent because it will be completely consumed before all the hydrogen is reacted.
What is the limiting reagent when 0.343 g of magnesium chloride is reacted with 10.0 mL of 2.00 M sodium hydroxide?
To determine the limiting reagent, calculate the number of moles of each reactant: 0.343 g of MgCl2 and 10.0 mL of 2.00 M NaOH. Convert 0.343 g of MgCl2 to moles and 10.0 mL of 2.00 M NaOH to moles to see which reactant is present in the lower stoichiometric amount. Whichever reactant yields the lower amount of product is the limiting reagent.
Can the mass of the limiting reagent be higher than the mass of the excess reagent?
it may be , the limiting reactant is that which is totally consumed during the reaction but its amount must be less than required amount with respect to excess reactant for example, H2SO4 + 2NaOH = Na2SO4 + 2H2O in this reaction suppose acid is 95 g and base is 85 g but acid with higher amount is the limiting reactant and base is in excess. Essentially, it's possible whenever the molecular weight of the limiting reagent is higher than the molecular weights of the other reagents.
30g of ammonium dichromate and 20 g of magnesium sulfate combined which will be the limiting factor?
Reactants: 30 g (NH4)2Cr2O7, 20 g MgSO4 30 g (NH4)2Cr2O7 / 252.07 g/mol = 0.119 mol 20 g MgSO4 / 120.41 g/mol = 0.166 mol Since they react in a 1:1 ratio to form magnesium dichromate and ammonium sulfate, ammonium dichromate is the limiting reagent (only 0.119 mol of MgSO4 is needed to react with all the (NH4)2Cr2O7).
Identify the limiting reagent and the volume of CO2 formed when 11 L CS2 reacts with 18 L O2 to produce CO2 gas and SO2 gas at STP. CS2(g) plus 3O2(g) and rarr CO2(g) plus 2SO2(g)?
To determine the limiting reagent, we need to compare the amount of CO2 that could be produced from each reactant. First, convert the volumes to moles using the ideal gas law. Then, use the coefficients from the balanced chemical equation to find the amount of CO2 that each reactant could produce. The limiting reagent is the one that produces the least amount of CO2. The volume of CO2 formed can be calculated using the stoichiometry of the limiting reagent.
If 0.25 g of Aluminum react with 0.6g of Copper sulfate then which one is the limiting reactant?
To determine the limiting reactant, we need to compare the moles of each reactant. First, calculate the moles of aluminum and copper sulfate separately. Then, determine the mole ratio between them and see which reactant is present in lower amount compared to the stoichiometric ratio. The reactant that is present in lower moles is the limiting reactant.
How do you know which reactant in a process is the limiting reactant?
The limiting reactant or reagent can be determined by calculating the number of moles of each reactant/reagent. Whichever is the lowest number of moles is the limiting reagent in the reaction, assuming that stoichiometry is 1;1
What is the limiting reactant in 2Al plus 3Cl2 equals 2AlCl3?
moles of Al=4.40 g/26.9815 g/mol=0.163 moles cl2=15.4g/70.906g/mol=0.217 the ratio is 2:3 cl2 is the limiting reagent
What is limiting reagent if 2.00 g of C2H4 are reacted with 5.00 g of O2?
1) balanced equationC2H4 + 2O2 --> 2CO2 + 2H2O2) convert mass to moles2.00 g C2H4 = 2.00/28.04= 0.0713 mol (3 significant figures)Referring to the balanced equation that means 0.143 mole of oxygen is required.5.00 g O2 = 5.00/32.00= 0.156 molThis is in excess of amount required. Therefore the ethene is the limiting reagent.
What is the limiting reagent when 150.0 g of nitrogen react with 32.1 g of hydrogen?
assuming you are talking about the synthesis of ammonia (NH3) N2 + 3 H2 --> 2 NH3 150.0 g / 28g/mole = 5.357 moles of nitrogen 32.1 g / 2.0g/mole = 16.05 mole of H2 5.357 moles N2 x 2 mol NH3 / 1 mol N2 = 10.714 mole ammonia (theoretical) 16.05 moles H2 x 2 mol NH3 / 3 mol H2 = 10.7 mole ammonia (theoretical) Almost the same, but 10.7 is slightly smaller. The smallest theoretical was found using hydrogen so hydrogen is the limiting reactant.
What is the answer and how did you get it using the balanced equation from problem 4 Determine the limiting reagent if you start with 200 grams of sodium hydroxide and 100 grams of sulfuric acid?
The chemical equation is:2 NaOH + H2SO4 = Na2SO4 + 2 H2OMolar mass of sodium hydroxide is 39,9971 g; molar mass of sulfuric acid is 98,079 g.2 . 39,9971 g NaOH----------------------98,079 g H2SO4200 g NaOH------------------------xx = (200 x 98,079)/2 . 39,9971 = 245 g H2SO4So sulfuric acid is the limiting reagent.
