1303f18exam1studentkey.pdf - Exam I CHEM 1303.001 FALL 2018 KEY(14 pts Calcium phosphide reacts with water to produce phosphine PH3 and calcium

1303f18exam1studentkey.pdf - Exam I CHEM 1303.001 FALL 2018...

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Exam I CHEM 1303.001 FALL 2018 KEY (14) pts. Calcium phosphide reacts with water to produce phosphine, PH 3 , and calcium hydroxide according the balanced equation below. Consider a reaction where 40.5 grams of PH 3 are produced, and 25.7 grams of Ca 3 P 2 remain unreacted. Needed molar masses for the problem are given below each formula. All table entries are in moles. Ca 3 P 2 ( s ) + 6 H 2 O( l ) à 2 PH 3 ( g ) + 3 Ca(OH) 2 ( s ) molar masses: 182 18.0 34.0 -- Initial (mol) 0.736 3.57 0 Change (mol) –0.595 –3.57 + 1.19 Final (mol) 0.141 0 1.19 a. Fill in all 9 table entries in moles . DO NOT CALCULATE AMOUNTS FOR Ca(OH) 2 . Show your work below for the entries above . At end of reaction: 40.5 g PH 3 1 mol PH 3 = 1.19 mol PH 3 34.0 g PH 3 25.7 g Ca 3 P 2 1 mol Ca 3 P 2 = 0.141 mol Ca 3 P 2 This is amount of excess reagent left over 182.0 g Ca 3 P 2 1.19 mol PH 3 6 mol H 2 O = 3.57 mol H 2 O This is amount of H 2 O that reacted; H 2 O is limiting. 2 mol PH 3 1.19 mol PH 3 1 mol Ca 3 P 2 = 0.595 mol Ca 3 P 2 This is amount of Ca 3 P 2 that reacted; Ca 3 P 2 is excess. 2 mol PH 3 (0.141 + 0.595) mol = 0.736 mol is the amount of Ca 3 P 2 present at beginning of reaction.
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