2 moles of br 27990 15980 g and that mass is 7995 of

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2 moles of Br = 2(79.90) = 159.80 g, and that mass is 79.95% of the total molar mass. So: MM = 159.80 0.7995 = 199.87 g/mol 199.87 g total – 159.80 g BR = 40.07 g = molar mass of element X. Looking in periodic table, the best match is Ca. (And note that CaBr 2 works as a compound because it includes Ca 2+ and Br ions.)
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A6 © 2015 L.S. Brown (10 pts) 5. Sodium bicarbonate (NaHCO 3 , also known as baking soda) can be used to neutralize acid spills. A lab accident causes two flasks to fall to the floor and break. One flask contained 125.0 mL of 1.20 M HNO 3 and the second contained 250.0 mL of 1.50 M H 2 SO 4 . Find the minimum mass of NaHCO 3 that should be used to completely neutralize the spilled acids. The relevant reactions are as follows. 2 NaHCO 3 (s) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + 2 CO 2 (g) + 2 H 2 O( ) NaHCO 3 (s) + NHO 3 (aq) NaNO 3 (aq) + CO 2 (g) + H 2 O( ) I will find the amount of NaHCO 3 needed to neutralize each acid and then add those together. For the HNO 3 : 0.125 L × 1.20 mol HNO 3 L × 1 mol NaHCO 3 1 mol HNO 3 = 0.150 mol NaHCO 3 For the H 2 SO 4 : 0.250 L × 1.50 mol H 2 SO 4 L × 2 mol NaHCO 3 1 mol H 2 SO 4 = 0.750 mol NaHCO 3 Add those to get 0.900 mol NaHCO 3 needed, and convert that to mass. 0.900 mol NaHCO 3 × 84.00 g NaHCO 3 1 mol NaHCO 3 = 75.6 g NaHCO 3
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NAME:_________________________________ © 2015 L.S. Brown A7 (12 pts) 6. Zinc sulfide reacts with oxygen according to the following equation. 2 ZnS(s) + 3 O 2 (g) 2 ZnO(s) + 2 SO 2 (g) A reaction mixture consisting of 312.6 g ZnS and 127.0 g O 2 is prepared and allowed to react, and 165.8 g of ZnO are produced. What was the percentage yield for this reaction? We know the actual yield, so to find the percent yield we need to get the theoretical yield first. To do that we need to work a limiting reactant calculation. We have: 312.6 g ZnS × 1 mol ZnS 97.46 g ZnS = 3.207 mol ZnS 127.0 g O 2 × 1 mol O 2 32 g O 2 = 3.969 mol O 2 So we can see that O 2 must be the LR.
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