Part D Enter your answers numerically separated by commas. mH SO,, mBaSO,, MHCI = Cubmit

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Part C
(2.61gBaCl2)+ (6.78GH,SO,)
- (rg BaSO4) + (yg HCI)
BaClh is the limiting reactant.
O H2SO4 is the limiting reactant.
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Since the reactants have a one to one stoichiometric relationship, the reactant with fewer moles will be the limiting reactant. Use the given mass of each
compound and divide each by its molar mass to find which reactant has fewer moles. For BaCl2,
mol BaCl,
= 2.61 g BaCl, ×
1 mol BaCl2
208.2 g BaCl2
= 0.0125 mol BaCl,
For H2SO4.
mol H2SO, = 6.78 g H2SO4 x
1 mol H2SO4
98.08 g H2SO4
0.0691 mol H2SO,
Thus, BaCla is the limiting reactant.
Part D
Enter your answers numerically separated by commas.
O Pearson
5:11 PM
59
5/2/2021
18
acer
Transcribed Image Text:Part C (2.61gBaCl2)+ (6.78GH,SO,) - (rg BaSO4) + (yg HCI) BaClh is the limiting reactant. O H2SO4 is the limiting reactant. Submit Previous Answers v Correct Since the reactants have a one to one stoichiometric relationship, the reactant with fewer moles will be the limiting reactant. Use the given mass of each compound and divide each by its molar mass to find which reactant has fewer moles. For BaCl2, mol BaCl, = 2.61 g BaCl, × 1 mol BaCl2 208.2 g BaCl2 = 0.0125 mol BaCl, For H2SO4. mol H2SO, = 6.78 g H2SO4 x 1 mol H2SO4 98.08 g H2SO4 0.0691 mol H2SO, Thus, BaCla is the limiting reactant. Part D Enter your answers numerically separated by commas. O Pearson 5:11 PM 59 5/2/2021 18 acer
= 0.0691 mol H2SO.
Thus, BaCl, is the limiting reactant.
Part D
Enter your answers numerically separated by commas.
MH,SO,, MBASO,, MHCI =
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Transcribed Image Text:= 0.0691 mol H2SO. Thus, BaCl, is the limiting reactant. Part D Enter your answers numerically separated by commas. MH,SO,, MBASO,, MHCI = Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining Provide Feedback Pearson
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