If a total of 13.5 mol of NaHCO_3 and 4.5 mol of C_6 H_8 O_7 react, how many mol

Question

If a total of 13.5 mol of NaHCO_3 and 4.5 mol of C_6 H_8 O_7 react, how many moles of CO_2 and Na will be produced? 3NaHCO_3 (aq) + C_6 H_8 O_7 (aq) rightarrow 3CO_2 (g) + 3H_2 (s) + Na The decomposition of potassium chlorate yields oxygen gas. If the yield is 95% how many grams of KClO_3 are needed to produced 15 g of O_2? 2KClO_3 (s) rightarrow 2KCl (s) + 3O_2 (g) If 5.0 g of H_2 are reacted with excess CO, how many grams of CH_3 OH are produced based on a yield of 86% CO(g) + 2H_2 (g) rightarrow CH_3 OH (l)

Explanation / Answer

Question 32

IF THE yield is 95% how many grams of kclo3 are needed to produce 15 g of o2

2KClO3(s]2KCl(s]+3O(g]

a 2 mole ratio between KClO3, and O2.

=15 * (1 mole O2/ 32 g) =0.468moles O2

=0.468moles O2 * (2/3) =0.312moles KClO3

95% =15 g/theoretical yield

theoretical yield =15 *100 /65 =23.07g

23.07g *(1 mole O2/32 )= 0.72moles O2

=0.72moles O2 *(2/3)= 0.48moles KClO3

=0.48moles KClO3 * (122.55 g/1 mole KCLLO3)= 58.824g KClO3

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