The compound P4S3 is used in matches. It reacts with oxygen to produce P4O10 and SO2. The unbalanced chemical equation is shown below. P4S3(s) + O2(g) → P4O10(s) + SO2(g) What mass of SO2 is produced from the combustion of 0.59 g P4S3? MM of P4S3 = 220.09 g.mol MM of SO2 = 64.07 g/mol Note: Round off your answer to two decimal places
The compound P4S3 is used in matches. It reacts with oxygen to produce P4O10 and SO2. The unbalanced chemical equation is shown below. P4S3(s) + O2(g) → P4O10(s) + SO2(g) What mass of SO2 is produced from the combustion of 0.59 g P4S3? MM of P4S3 = 220.09 g.mol MM of SO2 = 64.07 g/mol Note: Round off your answer to two decimal places
Chemistry for Engineering Students
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ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
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Chapter4: Stoichiometry
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Problem 4.111PAE: 4.111 Aluminum metal reacts with sulfuric acid to form hydrogen gas and aluminum sulfate (a) Write a...
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The compound P4S3 is used in matches. It reacts with oxygen to produce P4O10 and SO2. The unbalanced chemical equation is shown below.
P4S3(s) + O2(g) → P4O10(s) + SO2(g)
What mass of SO2 is produced from the combustion of 0.59 g P4S3?
MM of P4S3 = 220.09 g.mol
MM of SO2 = 64.07 g/mol
Note: Round off your answer to two decimal places
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