Relay Stoichiometry
Imagine the following reaction:
-- Na2SnO3 + H2S -> SnS2 + NaOH + H2O --
You are given 215 g of Na2SnO3 and 40 g of H2S.
Then you collect the H2O produced and use it in this reaction:
-- Ca3P2 + H2O -> Ca(OH)2 + PH3 --
You are given 175 g of Ca3P2. How many grams of PH3 is produced ideally?
Notes: -The equations above are NOT balanced. -Pick the choice that is the CLOSEST to your answer.
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1 solution
Step 1: Algebraically balance Na2SnO3 + H2S -> SnS2 + NaOH + H2O
aNa2SnO3 + bH2S -> cSnS2 + dNaOH + eH2O
Na: 2a = d Sn: a = c O: 3a = d + e H: 2b = d + 2e S: b = 2c
a = 1 b = 2 c = 1 d = 2 e = 1
Na2SnO3 + 2H2S -> SnS2 + 2NaOH + eH2O
Step 2: Find the limiting reagent of Na2SnO3 + 2H2S -> SnS2 + 2NaOH + eH2O:
215 g Na2SnO3 * 1 mol Na2SnO3/212.69 g Na2SnO3 = 1.01 mol Na2SnO3
40 g H2S * 1 mol H2S/34.076 g H2S = 1.17 mol H2S
H2S/Na2SnO3 = 2/1 1.17/1.01 = 1.15
H2S is the limiting reagent.
Step 3: Find H2O in mol.
1.17 mol H2S * 1 mol H2O/2 mol H2S = 0.585 mol H2O
Step 4: Balance Ca3P2 + H2O -> Ca(OH)2 + PH3
aCa3P2 + bH2O -> cCa(OH)2 + dPH3
Ca: 3a = c P: 2a = d H: 2b = 2c + 3d O: b = 2c
a = 1 b = 6 c = 3 d = 2
Ca3P2 + 6H2O -> 3Ca(OH)2 + 2PH3
Step 5: Find the limiting reagent of Ca3P2 + H2O -> Ca(OH)2 + PH3
175 g Ca3P2 * 1 mol Ca3P2/182.182 g Ca3P2 = 0.96 mol Ca3P2
H2O/Ca3P2 = 6/1 0.585/0.96 = 0.6
H2O is the limiting reagent.
Step 6: Find PH3 in grams.
0.585 mol H2O * 2 mol PH3/ 6 mol H2O * 33.998 g PH3 / 1 mol PH3 = 6.629 g PH3