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100.0 grams of sodium sulfate reacts with 50.00 grams of barium nitrate.   A-C IS ALREADY ANSWERED. ANSWER D-F QUESTIONS. a) What is the balanced equation? b) How many grams of barium sulfate will form? c) What is the limiting reactant? d) How much limiting reactant was used up in the reaction? e) How much excess reactant was used up for this reaction? f) How much of the excess reactant was left?

100.0 grams of sodium sulfate reacts with 50.00 grams of barium nitrate.   A-C IS ALREADY ANSWERED. ANSWER D-F QUESTIONS. a) What is the balanced equation? b) How many grams of barium sulfate will form? c) What is the limiting reactant? d) How much limiting reactant was used up in the reaction? e) How much excess reactant was used up for this reaction? f) How much of the excess reactant was left?

Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Chapter11: Stoichiometry
Section: Chapter Questions
Problem 75A
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Instruction: Final answer should be expressed with least number of significant figures.

Question: 100.0 grams of sodium sulfate reacts with 50.00 grams of barium nitrate.

 

A-C IS ALREADY ANSWERED. ANSWER D-F QUESTIONS.

a) What is the balanced equation?

b) How many grams of barium sulfate will form?

c) What is the limiting reactant?

d) How much limiting reactant was used up in the reaction?

e) How much excess reactant was used up for this reaction?

f) How much of the excess reactant was left?

needed . But herre there is 0 704 mol ,so Na,sOy, The baknced reachon is Na, so, + Ba (NO,), Na, SO4 Ba So, + a NANO, tor I mole of Na,SO, and Ba No,), , I mole of Basoy will form as it is char from balanced eachi'on moles are in 1o0q Na,so, many So have to find how we and Ba (NO,), Molar mass of Na, so, = 142 9/mol Molay mass of Ba Ng), = 261.34 g/mol the foomula Number of moles Can be hinel out uping given mass number of moles Molar mas Number of moles of Na, so4 1008 142g moi mol -1 0.704 Number of moles of Ba(Ng), 50 g 261.34 gmol" 0.191 mol We know that forr I mole Ba(NO,), I mole Na, soL o 191 mol of Ba(NO,), is available buf only so affer consuming o.191 BaNOg), no more will be left So only o191 mol of Na, so, will Alse we know that for I mole Ba (NO), mole ofc consumed. Also ,I mole of BaSOy will form.
Transcribed Image Text:needed . But herre there is 0 704 mol ,so Na,sOy, The baknced reachon is Na, so, + Ba (NO,), Na, SO4 Ba So, + a NANO, tor I mole of Na,SO, and Ba No,), , I mole of Basoy will form as it is char from balanced eachi'on moles are in 1o0q Na,so, many So have to find how we and Ba (NO,), Molar mass of Na, so, = 142 9/mol Molay mass of Ba Ng), = 261.34 g/mol the foomula Number of moles Can be hinel out uping given mass number of moles Molar mas Number of moles of Na, so4 1008 142g moi mol -1 0.704 Number of moles of Ba(Ng), 50 g 261.34 gmol" 0.191 mol We know that forr I mole Ba(NO,), I mole Na, soL o 191 mol of Ba(NO,), is available buf only so affer consuming o.191 BaNOg), no more will be left So only o191 mol of Na, so, will Alse we know that for I mole Ba (NO), mole ofc consumed. Also ,I mole of BaSOy will form.
That is for form. But we have to findd it in grams ০.1१। mo | ,0.191 mol of Basoy will numbenof moles of Baso, given molar mass of BasOy mass 0.191 mol given mass 233.38 glmol givenmas 4469 Roporting it to least number of signihicant Figure, 45 gof Baso, will form c) As Hgure out im s o 704 mol of Na, sa, and o191 molof above part we there is 0.704 BaNg), is available . So afer reaeling o191 mole 0.191 mole of Na,so, of Bu(NO)a with no more BalNO,), i available, so reachon will stop Liming reastant is the reaclant that get consumed competly in a reachion finst. So heve limiting neastant is BaNG),
Transcribed Image Text:That is for form. But we have to findd it in grams ০.1१। mo | ,0.191 mol of Basoy will numbenof moles of Baso, given molar mass of BasOy mass 0.191 mol given mass 233.38 glmol givenmas 4469 Roporting it to least number of signihicant Figure, 45 gof Baso, will form c) As Hgure out im s o 704 mol of Na, sa, and o191 molof above part we there is 0.704 BaNg), is available . So afer reaeling o191 mole 0.191 mole of Na,so, of Bu(NO)a with no more BalNO,), i available, so reachon will stop Liming reastant is the reaclant that get consumed competly in a reachion finst. So heve limiting neastant is BaNG),
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