How many grams of H2 will be produced from when 36 grams of HCI is allowed to react?

Md Ariful Islam

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Md Ariful Islam

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It is not clear with what substance HCL is allowed to react. However, from the common knowledge H2 will be formed upon reaction of HCl with metals, for example Mg in this case: Mg(s) + 2 HCl(aq) --> MgCl 2(aq) + H 2(g) Calculate number of moles of HCl: 36 g/36.5 g/mole =0.99 moles HCl According to the stoichiometry of the reaction number of moles of produced H2: 0.99 moles/2 = 0.49 moles, corresponds to grams oh H2 = 0.49 moles x 2 g/moles = 0.99 g H2 Note: the answer will be different if the metal reaction has different oxidation state. One need to write a balanced equation for each specific reaction
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CH4 + 4Cl2–>CCl4 + 4HCl if 9.32 grams of CH4 react with 2.71 grams Cl2 how many grams of CCl4 will be produced?

In first, You must to definite the excess substance in this case. Let's look to given reaction:CH4 + 4Cl2–>CCl4 + 4HClIt follows from it, that for 16 g CH4 need...

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Firstly, calculate the molar using the following equation M = 10XdensityXpercentage(37 in your case)/Molecular Weight. Then calculate the required molarity by dilution.

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When 15.25 grams of A reacts with 34.22 grams of B, 18.69 grams of C is produced. How many grams of D are produced?

30.78 g

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Ca(OH)2+2HCI=>CaCl2+H2O how many grams of H2O are produced when 86.4 grams of Ca(OH)2 reacts with 89.7 grams of HCI?

Molar mass of Ca(OH)2= 74.093 g/mol86.4/74.093= 1.166molMolar mass of HCl= 36.458 g/mol89.7/36.458= 2.46molMole ratio HCl: H2O= 2:1Mole ratio H2O:Ca(OH)2= 1:11.166+1.23=1.396molMolar mass of H2O = 18.0151.396×18.015= 25.15g

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25 mL x (1L/1000 mL) x (4 mol HCl/1L) x (1 mol Zn/2 mol HCl) x (65.38 g Zn/1 mol Zn) = 3.269g = ~ 3.3g of ZnParts explained25 mL...

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when heated, KCl3 decomposes into KCl and O2. 2KCl3= 2KCl, how many grams of O2 were produced? if this reaction produced 49.2g KCl, how many grams of O2 were produced?

2KClO3=2KCl+3O2Per 1 gram of KCl there are 0.644 g O2 produced.Per 49.2 gram of KCl there are 31.678 g O2 produced.

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If it requires 75.0 mL of 0.500M NaOH to neutralize 165.0 mL of an HCI solution, what is the concentration of the HCI solution?

Solution:Balanced chemical equation:NaOH + HCl → NaCl + H2OAccording to stoichiometry:n(NaOH) = n(HCl)CM(NaOH) × V(NaOH) = CM(HCl) × V(HCl)Therefore,CM(HCl) = CM(NaOH) × V(NaOH) / V(HCl)CM(HCl) = (0.500 M × 75.0...

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1) If Instead of using HCI, you used HNO3 with the same volume and concentration as HCI, would you use the same volume of NaOH to reach the endpoint? What about using H2SO4? Justify your answers

Actually, it strongly depends on what kind of concentration do you mean. If you mentioned molar concentration, then there is no deference between HCl and HNO3 in a reaction with...

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a). CuO(s) + 2HCI(aq) → CuCl2(s) + H2O(aq)b). Na2CO3(s) + 2HCl(aq) → H2O(aq) + CO2(g) + 2NaCl(aq)c). 3NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + 3H2O(aq)d). Ni(OH)2(aq) + 2CH3COOH(aq) → Ni(CH3COO)2(aq) +...

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calculate number of moles present in 20ml of hci solution to prepare a 0.02m hci acid

n=Cm x V = 0.02 x 0.02 = 0,0004 moles.

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